1st PUC

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Karnataka 1st PUC Kannada Textbook Answers Sahitya Sanchalana Chapter 2 Vachanagalu

Vachanagalu Questions and Answers, Notes, Summary

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Karnataka 1st PUC Chemistry Question Bank Chapter 11 The P-Block Elements

1st PUC Chemistry The P-Block Elements One Mark Questions and Answers

Question 1.
Why does carbon show maximum catenation among group 14 elements?
Answer:

1. Small size of carbon atoms.
2. C-C bond energy is comparable with the bond energy between carbon and other element
3. 4 valency electrons are present in four orbitals and the tetravalency of carbon is fully saturated.

Question 2.
Name different types of charcoal.
Answer:
Wood charcoal, Animal charcoal, Sugar charcoal.

Question 3.
Give one use of black diamond.
Answer:
Due to their hardness, black diamonds are used in various cutting tools.

Question 4.
What type of hybridization does carbon undergo in diamond and graphite?
Answer:
In diamond sp³, in graphite sp².

Question 5.
Silicon has become a vital element in the modern electronics industry. Why?
Answer:
It can function as semiconductors especially as n-type and p-type extrinsic semiconductors lubricants and insulators, Silicons are used for making water-proof papers by coating them with a thin layer of silicons.

Question 6.
What is catenation?
Answer:
Carbon atoms have a remarkable property of joining with one another in a large number to form a long chain and rings. This property is known as catenation or self linkage.

Question 7.
What is the technique used in preparing very pure silicon?
Answer:
Zone refining.

Question 8.
Give an example for intrinsic semiconductor other than silicon.
Answer:
Germanium in the pure state.

Question 9.
Who discovered fullerene 7
Answer:
Richard E. Smalley, Robert F. Curl Jr. and Harold W. Kroto.

Question 10.
What kind of semiconductor is obtained when a little of phosphorus is added to silicon?
Answer:
n- type semiconductor.

Question 11.
Name the oxidation states exhibited by Ge, Sn, and Pb.
Answer:
+2 and +4

Question 12.
What type of semiconductor is obtained by doping germanium with indium?
Answer:
p-type semiconductor

Questuon 13.
Mention the catenation capacity of Si.
Answer:
16.

Question 14.
Why is boron metalloid?
Answer:
Boron resembles both with metals and non-metals therefore, it is metalloid

Question 15.
Give two important ores of boron.
Answer:

1. Borax, Na₂B₄O₇10H₂O
2. Kemite, Na₂B₄O₇2H₂O

Question 16.
Name two isotopes of boron
Answer:
are isotopes of boron.

Question 17.
Name two allotropes of boron.
Answer:
Amorphous and crystalline.

Question 18.
Why does boron not form B3⁺ ion?
Answer:
Boron does not form B³⁺ ion due to smallest atomic size and highest ionization energy

Question 19.
Why is crystalline boron hard solid?
Answer:
It is due to strong covalent bonds.

Question 20.
Why is boron used in nuclear reactors?
Answer:
Bonor can absorb neutrons.

Question 21.
What is tincal ? Give its chemical formula.
Answer:
Tincal (or Borax) is an ore of boron. Its formula is NaB₄ Or 10H₂O

Question 22.
Why does boron resemble Si?
Answer:
Both have similar charge over radius ratio, i.e., similar polarizing power.

Question 23.
Why does boron form stable electron deficient compounds?
Answer:
Boron has three valence electrons, it will share three electrons with other elements to form electron deficient compounds which are stable.

Question 24.
Why is boric acid (H₃BO₃) monobasic acid?
Answer:
It accepts a pair of electrons from OH^– ion of H₂O therefore, it is monobasic acid.

Question 25.
What type of glass is obtained when borax is added ?
Answer:
Pyrex glass is a glass which is heat resistant. It can withstand high temperature.

Question 26.
What is use of diborane? Why BHs exists in form of diborane ?
Answer:
BH₃ exists in form of diborane (B₄H₆) because it is electron deficient. Diborane is used as reducing agent.

Question 27.
Complete the following
(i) B + O₂ →
(ii) B + N₂ →
(iii) B + Cl₂ →
(iv) BF3 + NH₂ →
(v) Na₂B₄O₇+H₂O →
(vi) Na₂B₄O₇ →
Answer:
(i) 4B + 3O₂ → 2B₂O₃

Question 28.
Why does \(\mathrm{BF}_{6}^{3-}\) not exist?
Answer:
It is because ‘B’ does not have vacant d-orbitals.

Question 29.
Which type of bonds are formed by boron and why?
Answer:
Boron forms covalent bonds because it cannot lose electrons or gain electron easily.

Question 30.
What is basic structural unit of orthoboric acid? Name type of bond present in them.
Answer:
\(\mathrm{BO}_{3}^{3-}\) is basic structural unit of boric acid. Covalent bonds are present.

Question 31.
Give structure of diborane.
Answer:

Question 32.
Why do boron halides form addition compounds with ammonia and amines?
Answer:
It is because they are electron deficient.

Question 33.
In which pure form does carbon exists in nature?
Answer:
Fullerene

Question 34.
Give structure of CO₂.
Answer:
O = C = O, It has linear structure.

Question 35.
What name is given to the compounds formed by more electropositive elements with carbon?
Answer:
Ionic compounds

Question 36.
Is carbon dioxide poisonous or not?
Answer:
No, it is not poisonous.

Question 37.
What is producer gas ?
Answer:
Producer gas is mixture of CO and N₂ in the ratio of 2 : 1

Question 38.
Buckminster fullerene is a crystalline allotrope of which element ?
Answer:
Carbon

Question 39.
What is state of hybridisation of C in \(\mathrm{co}_{3}^{3-}\) ?
Answer:
‘C’ in \(\mathrm{co}_{2}^{3-}\) has sp² hybridisation.

Question 40.
How does BF₃ act as catalyst in industrial process?
Answer:
BF₃ is electron deficient, therefore acts as catalyst in industrial processes.

Question 41.
Why is BF₃ weaker Lewise acid than BCl₃
Answer:
BF₃ is weaker Lewis acid than BCl₃ because of more effective bonding in case of F due to smaller size, than Cl.

Question 42.
Name the elements present in boranes.
Answer:
Boranes are made up of boron and hydrogen.

Question 43.
Write the state of hybridization in BF₃.
Answer:
‘B’ in BF₃ has sp² hybridisation

Question 44.
Why does boric acid act as Lewis acid ?
Answer:
It is because in boric acid, boron does not have its octet complete. It accepts OH- from
water in aqueous solution B(OH)₃ + H₂O → [B(OH)₄]^– + H⁺

Question 45.
Why does boron form electron deficient compounds ?
Answer:
It is because boron has three valence electrons. It shares three electrons and gets six electrons, i.e., its octet is not complete. Therefore, it forms electron deficient compounds.

Question 46.
Mention the chief reason for the anomalous behaviour of boron in group 13 of the periodic table.
Answer:
It is due to small size and higher ionization energy and high charge/size ratio of boron.

Question 47.
Mention the type of hybrid orbitals of silicon in \(\mathrm{SiF}_{6}^{2-}\) ion.
Answer:
sp³d²

Question 48.
Mention the state of hybridization of B in \(\mathrm{BH}_{4}^{-}\)
Answer:
sp³

Question 49.
How is that silicon atoms can have a co-ordination number more than four but carbon atoms cannot?
Answer:
Silicon has vacant d-orbitals, therefore, it can have coordination number more than four but carbon cannot have because it does not have vacant d-orbitals.

Question 50.
CO₂ is gas while SiO₄ is solid at room temperature. State a reason for this.
Answer:
CO₂ exists as decrete molecules, therefore, it has weak Van der Walls’ forces for attraction whereas SiO₂ is three dimensional covalent solid.

Question 51.
Between AIF₃ and AICl₃ which one will have a higher melting point.
Answer:
AIF₃ is more ionic, therefore, has higher melting point.

Question 52.
Why is that aluminium metal cannot be obtained by electrolysis of an aqueous solution of a salt of aluminium?
Answer:
It is because aluminium metal reacts with water at high temperature.

Question 53.
Which one of the following elements +1 oxidation state as well: Al, B, Ca, Tl, Be?
Answer:
TI shows +1 oxidation acid.

Question 54.
Carbon and Silicon are mainly tetravalent but Ge, Sn and Pb show divalency. Give reason.
Answer:
Ge, Sn, Pb are divalent due to inert pair effect which is not there in carbon and silicon.

Question 55.
What property of anhydrous AlCl₃ makes it a very good preparative reagent in organic chemistry?
Answer:
It acts as Lewis acid.

Question 56.
Although pure silicon is an insulator, then how does it behave as semiconductor on heating?
Answer:
Silicon becomes semiconductor on just heating because electrons becomes free to move.

Question 57.
What is name given to elements which are neither metals nor non-metals?
Answer:
Metalloids

Question 58.
What is general formula of silicones ?
Answer:
R₂SiO is general formula of silicones.

Question 59.
What is the oxidation state of Ni in [Ni(CO)₄]?
Answer:
Ni has oxidation state of zero in [Ni(CO)₄]

Question 60.
What is inert pair effect ?
Answer:
The pair of electrons in valence s-orbital is reluctant to take part in bond formation due to poor shielding effect of -d and f-electron in heavier elements. It is called inert pair effect due to which lower oxidation state becomes more stable than higher oxidation state in case of p-block elements.

Question 61.
How are linear silicones obtained ?
Answer:
Linear silicones are obtained by hydrolysis of R₂SiCl₂ (chlorosilanes)

Question 62.
In terms of atomic size, ionization energy and charge, explain why are ionic compounds containing B³⁺ not formed.
Answer:
Boron has smaller atomic size, therefore, it cannot lose three electrons because its third ionization energy is very high.

Question 63.
Why does C differ from rest of elements ?
Answer:
Carbon has smallest size, highest ionization energy and high electro negativity, therefore, it differs from rest of the elements.

Question 64.
What are silicates?
Answer:
Silicates are minerals which consist of \(\mathrm{SiO}_{4}^{4-}\) units arranged in different ways.

Question 65.
Write the formula of inorganic benzene.
Answer:
B₃N₃H₆

Question 66.
Explain the following statement with reason.
The fullerene is considered as purest allotrope of carbon.
Answer:
It is because it does not have edges, therefore, impurities cannot be absorbed on it.

Question 67.
Mention the nature of an aqueous solution of borax.
Answer:
Basic in nature.

Question 68.
Boric acid is polymeric. Why?
Answer:
The presence of hydrogen bonds makes it polymeric.

Question 69.
Mention the type of hybridization of boron in diborane.
Answer:
Sp³

Question 70.
Which is thermodynamically most stable form of carbon ?
Answer:
Graphite

Question 71.
State the trends observed in case of each of the following: Oxidation state of the elements of Group 14.
Answer:
Group 14 elements show +2 and +4 oxidation states. Tendency to show +2 oxidation state increases down the group due to inert pair effect.

Question 72.
How does the boric acid polymerise ?
Answer:
Through hydrogen bonding.

Question 73.
Does BH₃ exists?
Answer:
No.

Question 74.
How does boron interact with NaOH ?
Answer:
2B + 6NaOH → 2Na₃B0₃ +3H₂

Question 75.
What is the oxidation state of C in
(a) CO
(b) HCN
(c) H₂CO₃
(d) CaC₂.
Answer:
(a) +2, (b) +2, (c) +4, (d) -1

Questuon 76.
Give two examples of electron-deficient compounds.
Answer:
BF₃ and B₃H₆.

Question 77.
Arrange the following halides of boron in the increasing order of acidic character. BF₃, BCI₃, BBr₃, BI₃.
Answer:
BF₃ < BCI₃ < BBr₃ < BI₃.

Question 78.
Which form of carbon is used to decolourise sugar ?
Answer:
Bone charcoal.

Question 79.
What does we get when Cone. H₂SO₄ is dropped on sugar ?
Answer:
Sugar Charcoal.

Question 80.
What happens when a borax solution is acidified ? Write a balanced equation for the reaction.
Answer:
Boric acid is formed (Na₂B₃O₇ + 2HC1 + 5H₂)O →2Nacl + 4H₃BO₃

1st PUC Chemistry The P-Block Elements Two Marks Questions and Answers

Question 1.
Give the chemical reaction as an evidence for each of the following observation?

• Tin (II) is a reducing agent whereas lead (II) is not
• Gallium (I) undergoes disproportionation reaction.

Answer:

• It is because Pb²⁺ is more stable than Pb⁴⁺ due to inert pair effect whereas Sn⁴⁺ is more stable than Sn²⁺. Therefore Sn²⁺ is good reducing agent 3Ga⁺ —> 2Ga + Ga³⁺
• It is because Ga³⁺ is more stable than Ga⁺

Question 2.
Write the chemical equation for the reactions which occur when : R₂SiCl₂ is subjected to hydrolysis.
Answer:

Question 3.
Complete the following chemical reaction :
1. \(\mathrm{RCI}+\mathrm{Si} \frac{\text { Cupower }}{570 \mathrm{K}}\)
Answer:
2RC1 + Si → R₂SiCl₂

Questuon 4.
Describe two similarities and two dissimilarities between B and Al.
Answer:
Similarities :

1. Both have similar electronic configuration
2. Both have same number of valence electrons

Dissimilarities:

1. B is non-metal; Al is metal
2. B forms acidic oxide (B₂O₃) whereas Al form Al₂O₂ which is amphoteric

Question 5.
Draw a labelled diagram to show the dimeric structure of Al(III) chloride. State one use of anhydrous AICl₃.
Answer:
The structure of AICl₃ is shown below.

It issued in preparation of organic compounds in Friedal Crafts reaction.

Question 6.
Arrange the following compounds in decreasing order of property indicated against each. Give reason for your answer : BCI₃, AICl₃, GaGL₃, InCl₃, TiCl₃ (Stability of +3 oxidation state)
Answer:
BCl₃ > AICl₃> GaCl₃ > InCl₃ > TiCl₃ because of inert pair effect stability of +3 oxidation state decreases.

Question 7.
Discuss the diagonal relationship of Be and Al with regard to

• the action of alkali and
• the structure of the chlorides.

Answer:
Be and Al resemble with each other due to same charge over radius ratio.

• Both B and A1 amphoteric oxides
• Be and A1 form covalent compounds
• BeCl₂ exists as polymer whereas AICl₃ exists as dimmer
• Both Be and A1 react with NaOH to form similar compounds.

Question 8.
C and Si are always tetravalent but Ge, Sn, Pb show divalency.
Answer:
Ge, Sn, Pb show divalency due to inert pair effect, Pb²⁺ is more stable than Pb⁴⁺.

Question 9.
Which of the following is acidic and why: SiCO₂, AI₂O₃, PbCO₂, SiCO₂?
Answer:
SiCO₂ is acidic oxide because Si is non-metal.

Question 10.
As we move down in group 13 elements increase in atomic size is comparatively very less. Explain.
Answer:
It is due to poor shielding effect of d and f-electrons, as result of which effective nuclear charge increase.

Question 11.
Write the chemistry involved in etching of glass.
SiO₂ + 4HF → SiF₄ + 2H₂O
Answer:
SiF₄ +2HF → H₂SiF₆(Fluorosilicicacid)

Question 12.
Elemental silicon does not form graphite like structure as carbon does. Give reason.
Answer:
It is because ‘Si’ cannot form pπ-pπ bond like carbon in graphite due to larger atomic size.

Question 13.
What are silicones ?
Answer:
Silicones are rubber like polymers.

Question 14.
Give the shapes of
(a) SiF4,
(b) \(\mathrm{SiF}_{6}^{2-}\)
Answer:

Questuon 15.
Arrange the following in decreasing order of boiling point giving reason CH₄, GeH₄, SnH₄, SiH₄
Answer:
SnH₄ > GeH₄ > SiH₄ > CH₄
As we go down the group from C to Sn surface area of hydrides increases therefore, Van der Waals’ force increases hence boiling point increases.

Question 16.
Explain the following: (a) CaO reacts with SiCO₂ to form a slag.
Answer:
(a)

1st PUC Chemistry The P-Block Elements Threer / Four / Five Marks Questions and Answers

Question 1.
What are allotropy ? Mention allotropy of Pb, Sn, Ge, Si.
Answer:
Allotropy: The existence of same element in two or more forms in the same physical state but having different properties called allotropy. The different forms are called allotrope.
Pb : —> No allotrope exists.
Sn : —> Grey ( ) = White ( ) = ; Rhombic { Tin Cry : the sound when crystals of tin is Bent or deformed}
Ge : —> Two (Both are crystalline form.)
Si: —> Two ( Both are crystalline form.)

Question 2.
Explain the oxidation property of group 14 elements.
Answer:
Group 14 elements show commonly +2 and +4 oxidation state. The carbon family also shows both positive and negative oxidation state. It shows +2 oxidation state when np² electrons involved in the bond formation Ex-SnCl₂, PbSO₄. This show +4 oxidation state when ns²np² electrons involved in the bond formation. Ex-SnCl₄

Question 3.
Explain semiconductors with its types and uses.
Answer:
Semiconductors are the materials their conductance is lies between those of conductors and insulators. They allow only a certain portion of the current to pass through them. Materials used for semiconductors should be extremely pure and is obtained by zone refining.

Question 4.
Explain the types of semiconductors.
Answer:
1. Intrinsic Semiconductors: If Si of Ge is sufficiently heated the energy supplied breaks some of the covalent bonds and electrons become free to carry current. When the electrons are get free are released, positive holes are left behind in the lattice. In intrinsic semiconductors, the number of holes and electrons will be equal.

2. Extrinsic semiconductors: These are formed doping (addition of impurities of certain elements) to the insulators.

Question 5.
Explain the types Of Extrinsic Semiconductors
Answer:
I. n-type semiconductor: Doping of (Group 14 elements) Si, Ge, with pentavalent impurities (group 15 elements) like P. As, produces excess of electron and they are called n-type semiconductors. The symbol ‘n’ indicates the flow of negative charge in them.

Question 6.
How does metallic character vary among the group 14 elements? How is it related to ionisation energy?
Answer:
Carbon and silicon are non metals. Germanium is a metalloid and tin and lead are metals. Metallic character increases from carbon to lead. The change from non metallic to metallic character is due to the increase in atomic size from carbon to lead. Consequently the ionization potential decreases from carbon to lead.

Hence the electropositive character increases favouring ionic bonding especially with oxidation state +2. Whereas carbon and silicon show +4 oxidation state forming covalent bonding with electron sharing.

Question 7.
Distinguish between n- type and p – type semiconductors.
Answer:
n – type semiconductors are formed by doping pure Si or Ge with pentavalent elements like P, As, Sb, etc, Of the five valence electrons of the added elements, four of them form covalent bonds with Si (or Ge). The fifth electron, conducts the current.

In p – type conductor Si or Ge is doped with trivalent elements like Boron, aluminum or indium. The three valence electron of the added element form three covalent bonds with Si (or Ge) results as deficiency of one electron in the added element causes a positive hole. This type of extrinsic semiconductors are known as p – type semiconductors ( p – stands for positive hole).

Question 8.
How do you account for the tetravalency of carbon even though there are only 2 unpaired electrons ?
Answer:
Electronic configurations of carbon atom in the ground state is \(1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{P}_{2}^{1} 2 \mathrm{P}_{2}^{1} 2 \mathrm{P}_{2}^{0}\) During the formation of a compound the activation energy of the reactants excites the electrons to give the configuration as \(1 \mathrm{s}^{2} 2 \mathrm{s}^{1} 2 \mathrm{p}_{\mathrm{x}}^{1} 2 \mathrm{p}_{\mathrm{y}}^{1} 2 \mathrm{p}_{\mathrm{z}}^{1}\) Then four orbitals each containing an impaired electron, mix together and form four hybridized sp³ orbitals directed from the carbon atom tetrahedrally. Hence carbon exhibits tetravalency.

Question 9.
How do the conductivity or semic conduction can be increased?
Answer:
As the temperature increases, the number of covalent bonds ruptured in an intrinsic semiconductor increases. This results in the increase of the electron hole pairs. As a result, the conductivity of the intrinsic semiconductor increases with rise in temperature. When electron field is applied across these materials electron migrate in one direction and the positive holes migrate is the opposite direction and thus the material conducts. At a given temperature the conductivity of an intrinsic semiconductor is constant.

Conductivity of a semiconductor can be increased by adding either pentavalent or trivalent elements. The elements added to intrinsic semiconductors to increase conductivity are called dopants, the process of addition of dopants (impurities) is called doping. The dopped semiconductors are called extrensic Semiconductors.

Question 10.
Write a note on structure of Fullerene. Mention any two of its application.

Answer:
Richard E. Smalley; Robert F. Curl Jr and Sir Harold W. Kroto the discoverers of this new form of pure carbon laked fullerene, they shared the 1996 Nobel Prize in chemistry for their discovery.

This third crystalline form of carbon discovered was named as buckminster fullerene which is commonly called fullerenes. They were named so because their structure resembles geodestic domes designed by the American scientist, engineer and philosopher R. Buckminster Fuller, They are also philosopher R. Buckminster Fuller. They are also nicknamed “Bucky ball” because their shape resembles a soccer ball.

Fullerenes are large spheroidal C_n molecules with n = 60 or more. When an electric arc is struck between graphite electrodes in an inert atmosphere a large quantity of soot is formed together with a significant quantity of C₆₀ and much smaller quantities of inter related fullerenes such as C₇₀ ; C₇₆ ; C₇₈.

When graphite is vapourised using a high power laser, fullerene C₆₀ is obtained. The fullerene C₆₀ bucky ball, has the same form as in soccer ball. In this each carbon atom is sp² hybridised. Each carbon atom has three neighbouring carbon atoms and forms two single and a double bond. C₆₀ has 32 faces. 20 of which are hexagons and 12 of which are pentagons. Each pentagon is surrounded by five hexagons and each hexagon is linked to three pentagons and each three hexagons. It has 60 vertices with a carbon atom at each vertex.

Question 11.
Write the characteristic uses of fullerene.
Answer:
C₆₀ dissolves in hydrocarbon solvents such as hexane, benzene, toluene etc giving magenta solutions. Fullerenes are the least stable of the carbon allotropes and graphite is the most stable. Fullerenes react with alkali metals to produce solids having composition such as K₃ C₆₀. This compound acts as a superconductor below 18K. Fullerene C₁₁₄ has more surface area and can withstand a very high temperature. Hence it can be used as a lubricant in satellites. Fullerenes are used as industrial catalyst and as lubricant in the treatment of cancer.

Question 12.
Explain the structure of graphite giving the reason for its softness as well as its electrical conductivity.
Answer:
Good conductor of electricity : In. graphite carbon atom is sp² hybridised. Out of four valiancy electrons in each carbon atom only three electrons, one from 2s and two from 2p are involved in hybridisation. One spare electron in its 2p_z orbital is left free. These 2p_z orbital overlap to form delocalised π system which extends above and below each layer. In this delocalised π electrons are free to move within the layer and hence graphite is a good conductor of electricity.

The conductivity of graphite perpendicular to the plane of the layers of hexagons is low and increases with increase in temperature, signifying that graphite is a semiconductor in that direction. The electrical conductivity is much higher parallel to the plane but decreases as the temperature is raised.

Soft, flaky and slippery substance : In graphite each layers of hexagon are held together by a weak Vander Waal’s force of attraction at a large distance of 3.40 Å. Even the slight pressure causes the layers to slide over one another. Hence graphite is soft, flaky and slippery. Graphite is used as a lubricant.

Question 13.
Explain the structure of diamond giving the reason for its hardness as well as non – conductivity of electricity.
Answer:
Non-conductor of electricity: All the four electrons present in the valence shell of carbon atoms are used up in sp³ hybridisation and form sigma bonds. No mobile electrons are left in the diamond crystals to allow the conduction of electricity. Hence diamond is a non-conductor of electricity.

Extremely hard: Due to high strength of covalent bonds holding all the carbon atoms in diamond together, diamond is very hard. Infact, diamond is the hardest substance known.

Question 14.
Give the difference in structure of the following pair of compounds : CO₂ and SiO₂.
Answer:
CO₂ linear molecule and exists as monomer. It is gas while SiO₂ is solid at room temperature due to three dimensional network in which each Si atom is covalently bonded to four oxygen atoms tetrahedrally. In CO₂, ‘C’ is sp-hybridised while in SiO₂, ‘Si’ is sp³ hybridised. CO₂ is discrete molecule where as SiO₂ is 3-dimensional solid.

Question 15.
Explain the formation of

(i) water gas
(ii) producer gas. Give their uses.
Answer:

Water gas and producer gas are used as fuel

Question 16.
What happens when CO₂ is passed through lime water
(i) for short duration
(ii) for a long duration ?
Answer:

Question 17.
The first element in every group of representative elements shows properties different from the characteristic properties of the group. Name three such elements and mention two abnormal properties of each one of them.
Answer:
Boron carbon Nitrogen
Boron

• W, forms acuhc. oxides -whereas owners item. am-phofork and.bask, oxide.
• It cannot form [BFe]^3′ whereas others can form such type of complexes.

Carbon:

• It shows property of catenation to maximum extent.

Nitrogen

• N₂ is gas others are solid
• NH₂ is liquid, other hydrides are gases.

Question 18.
Give equations for the following:

• Preparation of CO₂,
• Basic properties of Na₂CO₃,
• Formation of acetylene.

Answer:

• CaCO₃ + 2HC1 —> CaCl₂ + CO₂ + H₂O
• Na₂CO₃ + 2H₂0 —> 2Na0H + H₂CO₃
• CaC₂(calcium carbide) + 2H₂0 —>Ca(OH)₂ + C₂H₂(Acetylene)

Question 19.
What is dry ice ? Why is it called so? How will you prepare pure sample of CO (Carbon monoxide)?
Answer:
Solid CO₂ is called dry ice. It is called dry ice because it directly changes into gaseous state without becoming liquid.

Question 20.
What happens when ?

• Quick lime is heated with coke ?
• Carbon monoxide reacts with Cl₂?
• Plants absorb CO₂?

Answer:

• CaO + 3C —> CaC₂(CalciumCarbide) + CO,
• CO + Cl₂ —> COCl₂ (Phosgene)
• 6CO₂ + 6H₂O —> C₆H₁₂O₆(glucose) + 6O₂

Question 21.
Write balanced equations for the following reactions and name the main product formed in each case,

• NaBH₄ +I₂ —>
• B₂h₆ +NaH —>
• \(\mathrm{BF}_{3}+\mathrm{LiH} \stackrel{450 \mathrm{K}}{\longrightarrow}\)
• SiCl₄ + H₂O —>

Answer:

• 2NaBH₄ +I₂ —> 2Nal + +H₂ + B₂H₆ (Diborane)
• B₂H₆+2NaH —> 2NaBH₄ (sodium borohydride)
• 2BF₃ + 6LiH —> 6LiF + B₂H₆ (Diborane)
• SiCl₄ + 2H₂O —> 4HCl + Si(OH)₄ (Silicicacid)

Question 22.
Give a comparative account of the chemistry of carbon and silicon with regard to their

• property of catenation and
• stability of hybrids and oxides.

Answer:

• Carbon shows property of catenation to more extent than silicon due to small size and tendency to form pπ-pπ multiple bonds with itself
• Stability of hybrids : CH₄ is more stable than SiH4 due to small size of ‘C’.
• CO₄ is gas whereas SiO₂ is three dimensional covalent solid, therefore highly stable as compared to CO₂.

Question 23.
A certain salt X, gives the following results.
(i) Its aqueous solution shows alkaline to litmus paper
(ii) It swells up to a glassy material Y on strong heating
(iii) When cone H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates out-Write equations for all the above reactions and identify X, Y and Z.
Answer:

Question 24.
Write balanced equations for:

1. BF₃ + LiH →
2. B₂H₆ + H₂O →
3. NaH + B₂H₆ →
4. \(\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{\Delta}{\longrightarrow}\)
5. A1 + NaOH →
6. B₂H₆ + NH₃ →

Answer:

1. 2BF₃ +6LiH →B₂H₆ +6LiF

2. B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂

3. 2NaH + B₂H₆ → 2NaBH₄

4. \(\begin{array}{l}{\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{HBO}_{2}+\mathrm{H}_{2} \mathrm{O}} \\ {4 \mathrm{HBO}_{4} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{B}_{2} \mathrm{O}_{3}+2 \mathrm{H}_{2} \mathrm{O}}\end{array}\)

5. Al + 3NaOH → A1(0H)₃ +3Na

6. \(3 \mathrm{B}_{2} \mathrm{H}_{6}+6 \mathrm{NH}_{3} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}+12 \mathrm{H}_{2}\)

Question 25.
Give one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
Answer:

Question 26.
(a) Carbon dioxide is non-polar while water is polar. What conclusion do you draw about their structures from these.
(b) Classify the following compounds into acidic, basic and amphoteric oxides. Al₂O₃, Cl₂O₇
Answer:
is linear, bond moments are equal and opposite, net dipole moment is zero. Water is bent molecule, it has net dipole moment .

Al₂O₃ is amphoteric where Cl₂O₇ is acidic oxide.

Question 27.
A white crystalline solid ‘A’ dissolves in water to give an alkaline solution. On heating ‘A’ first loses water molecules and swells up. On further heating it turns into a transparent liquid which solidifies into a glassy bead.
Answer:
‘A’ is Borax i.e. Na₂[B₄O₅(OH)₄].8H₂O which is also written as.
The reactions are:

Question 28.
Boric acid is a weak acid. It acts as an acid only on first receiving OH- from the medium and then releasing H+ ions. From the following organic compounds; select the one in presence of which it acts as a strong acid. Glycerol, Ethylene, Ethylalcohol, Acetic Acid.
Answer:
It acts as a strong acid in presence of glyceol.
Reason: Acetic acid itself is an acid and cannot give OH^– ion, ethyl alcohol being a primary alcohol is alos acidic. Ethylene also can’t furnish OH^– ion. But glycerol that has 3-OH groups. One of which is a secondary -OH group can assist boric acid to release H+ ions.

Question 29.
A compound ‘X’ or reduction with LiAH4 gives a hydride ‘Y that contains 21.72% hydrogen along with other products. The compound ‘Y’ reacts with air explosively resulting in boron trioxide. Identify ‘X’ and ‘Y’. Give balanced reactions involved in the formation of ‘Y’ and its reactions with air. Draw the structure of Y.
Answer:
“X’ is BCl₃ Boron Trichloride Y is B₂H₆ as percentage of hydrogen given is 21.72%

Question 30.
Mention three important uses of borax.
Answer: It is used :

• as a flux soldering and welding industry
• in the manufacture of borosilicate glass (or pyrex glass)
• in making enamels and glazes
• in stiffening of candle wicks
• in softening of water
• in qualitative analysis for borax bead test in laboratory.

Question 31.
Mention some important properties of carbon monoxide.
Answer:
1. It is colourless, odourless gas, slightly soluble in water.

2. It is highly poisonous. It combines with haemoglobin in the red blood cells to form carboxy-haemoglobin which cannot absorb oxygen and thus supply of oxygen to the body is reduced.

3. It burns with a pale blue flame forming CO₂; i.e., 2CO + O₂ → 2CO₂

4. It is a reducing-agent. It reduced some metal oxides into metal,
Fe₂O₃ + 3CO → 2Fe + 3CO₂

5. It combines with transition metals like iron, cobalt, nickel to form their carbonyl compounds;

Question 32.
What are halides of carbon ? Give few examples.
Answer:
Carbon combines with halogens to form both simple and mixed tetrahalides. In case of simple halides all the four expected tetrahalides (e.g. CF₄, Cl₄, CBr₄ and Cl₄) are known to exist. The stability of the simple tetrahalides (CF₄ > CCl₄ > CBr₄ > Cl₄)
Amongst the mixed halides the better known compounds are
(CFCl₃,CF₂Cl₂ and CCl₃Br)

Question 33.
What is allotropy ? Give examples of allotropes.
Answer:
Two or more forms of the same elements in same physical state which differ in their physical properties but have same chemical properties are called allotropic forms or (allotropes) and the phenomenon is called allotropy.
Carbon, phosphorus and sulphur are some elements which exhibit allotropy.

• Diamond and graphite are allotropic forms of carbon
• Red phosphorus and white phosphorus are allotropes of phosphorus.
• Rhombic sulphur, monoclinic sulphur and plastic sulphur are allotropic forms of sulphur.

Question 34.
Draw the structure of
(a) anion cyclic silicates
(b) anion sheet silicates.
Answer:

Question 35.
Give three uses of different allotropic forms of carbon.
Answer:

Forms of carbon	Uses
Diamond	Gemstone, cutting, drilling, grinding, polishing, industry.
Graphite	Reducing agent, refractories, pencils, high temperature, crucibles, electrode making, moderator in nuclear reactors, high strength composite materials.
Activated carbon	Rubber industry, pigments in ink, paints and plastics
Coke	Fuel, strut manufacture
Charcoal	Fuel, reducing agent, Adsorption.

Question 36.
Write balanced equation for the following:
(i) BF₃ + LiH →
(ii) B₂H₆+H₂O →
(iii) NaH + B₂H₆ →
(iv) H₃BO₃ →
(v) A1 + NaOH + H₂O →
(vi) B₂H6 + NH₃ →
Answer:

Question 37.
Draw the structure of fullerene C₆₀
Answer:

Question 38.
Draw the structure of graphite.
Answer:

Question 39.
How does electron deficient compound BF₃ achieve electronic saturation i.e., the fully occupied outer electron shells ?
Answer:
1. By multiple bonding, e.g., BF3 in which a lone pair of electrons of each fluorine atom may be used in a B → F π-bond (back bonding) involving the vacant orbital on the boron atom.

2. By formation of complexes in which electrons are received from a donor molecule, e.g., F₃B → NH₃. Boron compounds, thus, behave as Lewis acids.

Question 40.
Compound X on reduction with LiAH4 gives a hydride Y containing 21.72% hydrogen along with other products. The compound Y reacts with air explosively resulting in boron trioxide. Identify X and Y. Give balanced equations involved in the formation of Y and its reaction with air. Draw the structure of Y.
Answer:
Step 1: To determine the molecular formula and structure of compound Y.
1. Since the hydride Y reacts with air forming boron trioxide, therefore, Y must be an hydride of boron.
2. %H = 21.72% (Given)

∴ Emperical formula of Y = BH₃
Since boron forms two types of hydrides, i.e., B_nH_n+4 (nidoboranes) and B_nH_n+6 (arachnoboranes), therefore, Y must be a nidoborane with n = 2. Thus, M.F of Y = B₂H₆ . If Y is B₂H₆ (diborane), then its structure must be as follows :

Bridges B…H = 134 pm Terminal B – H = 119 pm

Step 2: To determine the structure of the compound X.
Since compound Y i.e., B₂H₆ is formed by reduction of compound X with LiAH₄ ,
therefore, X must either BCl₃ or BF₃.

The equation representing the reaction of Y with O2 may be written as follows :
B₂H₆ + 3O₂ → 2B₂O₃ + 3H₂O
Diborane,Y
Thus X = BF₃ and Y = B₂H₆

Question 41.
How is boron obtained from boran? Give chemical equations with reaction conditions and its reaction with HCl₁
Answer:

Question 42.
Which one is more soluble in diethyl ether, anhydrous AlCl₃ or hydrous AICl₃ ? Explain in terms of bonding ?
Answer: An hydrous AICI₃ is an electron-deficient compound while hydrated AICI₃ is not. Therefore, anhyd. AICI₃ is more soluble in diethyl ether because the oxygen atom of either donates a pair or electrons to the vacant p-orbital on the A1 atom in AICI₃ forming a coordinate bond.

In case of hydrated AlCl₃. Al is not electron deficient since H₂O has already donated a pair of electrons to it.

Question 43.
Mention the structures of (CH₃)₃N and (Me₃Si)₃ N. Are they isostructural? Justify your answer.
Answer: (CH₃)₃ N is pyramidal involving sp³. hybridisation of N atom. Nitrogen in (Me₃Si)₃ is however, sp² -hybridized. The reason being that the p-orbit of N containing the lone pair of electrons overlaps with a vacant d-orbital of silicon. As a result of this pπ – dπ back bonding, (Me₃Si)₃ N is planar.

Question 44.
1. Describe with equation what happens when carbon is heated with cone. H₂SO₄ ?
2. State how will you separate C0₂ and S0₂ from the mixture?
Answer:
1. H₂S0₄ oxidies C to CO₂ and itself is reduced to SO₂

2. Since the O.N. of C is CO₂ is +4 which is the maximum for C, therefore, it cannot act as a reducing agent. In contrast the O.N. of S is SO₂ is also +4 but its maximum. O.N. +6, therefore, it can act as reducing agent. This reducing gas can be removed by passing the mixture of these gases through acidified K₂Cr₂O₇ solution when orange solution turns green due to reduction of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ion to Cr³⁺ ions while CO₂ passes out unreacted.

Question 45.
What is inorganic benzene ? Why is it so called ? How will you get it from diborane?
Answer:
Borazine or borazole (B₃N₃H₆) is known as inorganic benzene. It is also called Because the structure of borazine is similar to that of benzene.
It is also isoelectronic and isosteric with benzene.

Like carbon is benzene, both N and B in borazine are sp² hybridized. Each N has a p-orbtal which is perpendicular to the a-bonding orbitals and contains a lone pair of electrons. In contrast, each B has an empty p-orbitals which is also perpendicular to the plane of the ring. Thus, the x-bonding in borazine is dative and it arises from the sideways overlap of fully filled orbitals of N and empty p-orbitals of B.

Preparation of borazine : At low temperature, diborane combines with ammonia to form an addition compound.
\(\mathrm{B}_{2} \mathrm{H}_{6}+2 \mathrm{NH}_{3} \stackrel{\text { Low Temp. }}{\longrightarrow} \mathrm{B}_{2} \mathrm{H}_{6} \cdot 2 \mathrm{NH}_{3}\)
When this addition compound is heated to 473 K, it decomposes to form a volatile compound called borazine.

Question 46.
State with equations what happens when borax is heated on a platinum wire loop and to the resulting transparent mass a minute amount of CuO is added and the mixture is again heated first in the oxidizing flame and then in the reducing flame of a Bunsen Burner ?
Answer:
When borax is heated on a platinum where loop, a transparent glassy bead is formed

When this transparent glassy bead is heated with CuO in the oxidizing flame, it imparts blue colour to the bead due to the formation of cupric metaborate
\(\mathrm{CuO}+\mathrm{B}_{2} \mathrm{O}_{3} \frac{\text { oxidising }}{\text { flame }} \mathrm{Cu}\left(\mathrm{BO}_{2}\right)_{2}\)(Cupric metaborate (blue)
However, when cupric metaborate is heated in the reducing flame of the Bunsen burner, the blue cupric metaborate is reduced by carbon present in the flame first to colourless cuprous metaborate and then to metal which produces red colour.

You can Download Chapter 21 Neural Control and Coordination Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Question Bank Chapter 21 Neural Control and Coordination

1st PUC Biology Neural Control and Coordination NCERT Text Book Questions and Answers

Question 1.
Briefly describe the structure of the following:
(a) Brain
(b) Eye
(c) Ear
Answer:
(a) Brain
The human brain is well protected by the skull. Inside the skull, the brain is covered by cranial meninges consisting of three layers namely outer durometer, middle layer called arachnoid and inner layer pia mater. The brain is divided into three major parts.

• Forebrain
• Mid brain
• Hindbrain.

Forebrain:
The forebrain consists of cerebrum, thalamus and hypothalamus. Cerebrum is divided longitudinally into two halves, left and right cerebral hemispheres. The hemispheres are connected by a tract of nerve fibres called corpus callosum. The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. It is referred to as grey matter due to greyish appearance. The cerebral cortex contains motor areas, sensory areas and association areas which are responsible for complex functions like intersensory associations, memory and communication.

Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. This layer is white in colour, hence called white matter. The cerebrum wraps around a structure called thalamus, which is a major co-ordinating centre for sensory and motor signalling. Hypothalamus lies at the base of the thalamus and contains several group of neurosecretory cells, which secrete hormones called hypothalamic hormones. The inner parts of cerebral hemispheres and a group of associated deep structures like amygdala, hippocampus, etc. form a complex structure called the limbic lobe or limbic system.

(ii) Mid brain:
The midbrain is located between the thalamus / hypothalamus of the forebrain and pons of the hind brain. A canal called the cerebral aqueduct passes through the mid brain. The dorsal portion of the midbrain consists of four round swellings (lobes) called Corpora quadrigemina. Midbrain and hindbrain form the brain stem.

(iii) Hindbrain:
Hindbrain comprises pons, cerebellum and medulla. Pons consists of fibre tracts that interconnect different regions of the brain. Cerebellum has very convoluted surface in order to provide the additional space for many more neurons. The medulla of the brain is connected to the spinal cord.

(b) Eye:

The adult human eye ball is nearly a spherical structure. The wall of the eye ball is composed of three layers. The external layer called Sclera is composed of a dense connective tissue. The middle layer choroid contains many blood vessels and looks bluish in colour. The choroid layer is thin over the posterior two – thirds of the eyeball, but it becomes thick in the anterior part to form the ciliary body. The ciliary body continues to form a pigmented and opaque structure called the iris.

The eyeball contains a transparent crystalline lens which is held in place by ligaments attached to the ciliary body. The aperture surrounded by the iris is called the pupil. The diameter of the pupil is regulated by the muscle fibre of iris. The inner layer is the retina and it contains three layers of cells called ganglion cells, bipolar cells and photoreceptor cells. Rods and cones are two types of photoreceptor cells. The optic nerves leave the eye and the retinal blood vessels enter it at a point medial to and slightly above the posterior pole of the eye ball. Photoreceptor cells are not present in that region and hence it is called the blind spot.

At the posterior pole of the eye lateral to the blind spot, there is a yellowish pigmented spot called macula lutea with a central pit called the fovea. The fovea is a thinned out portion of the retina where only the cones are densely packed. It is the point where the visual activity is the greatest.

The space between the cornea and the lens is called the aqueous chamber and contains a thin watery fluid called aqueous humor. The space between the lens and the retina is called the vitreous chamber and is filled with a transparent gel called vitreous humor.

(c) Ear
Ear is divided into three major sections called the outer ear, the middle ear and the inner ear. The outer ear consists of the pinna and external auditory meatus. The external auditory meatus leads inwards and extends up to the tympanic membrane (ear drum). There are very fine hairs and wax- secreting sebaceous glands in the skin of the pinna and the meatus. The tympanic membrane is composed of connective tissues.

Covered with skin outside and with mucus membrane inside. The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. An Eustachian tube connects the middle ear cavity with the pharynx.

The fluid filled inner ear called labyrinth consists of two parts, the bony and the membranous labyrinths. The bony labyrinth is a series of channels. Inside these channels lies the membranous labyrinth, which is surrounded by a fluid called perilymph. The membranous labyrinth is filled with a fluid called endolymph. The coiled portion of the labyrinth is called cochlea.

The membranes consisting cochlea, the reissners and basilar, divide the surrounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibuli ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear.

The organ of Corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti. The basal end of the hair ceil is in close contact with the afferent nerve fibres. A large number of processes called stereo cilia are projected from the apical part of each hair cell.

Above the rows of the hair cells is a thin elastic membrane called tectorial membrane. The inner ear also contains a complex system called vestibular apparatus, located above the cochlea. The vestibular apparatus is composed of three semicircular canals and the otolith organ consisting of the saccule and utricle. Each semi circular canal lies in a different plane at right angles to each other. The membranous canals are suspended in the perilyrrph of the bony canals. The base of canals is swollen and is called ampulla, which contains a projecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a projecting ridge called macula.

Question 2.
Compare the following:
(a) Central neural system (CNS) and Peripheral neural system (PNS)
(b) Resting potential and action potential
(c) Choroid and retina
Answer:
(a) The central neural system (CNS) includes the brain and the spinal cord and is the – site of information processing and control. The PNS comprises of all the nerves of the body associated with the CNS (brain and spinal cord)

(b) Resting potential is the electrical potential difference across the resting plasma membrane. Here the outer surface of the axonal membrane is positively charged and inner surface is negatively charged and therefore is polarised.

The electrical potential difference across the plasma membrane, when the membrane is depolarised is called the action potential. Here the outer membrane is negatively charged and inner membrane is positively charged.

(c) Choroid layer is the middle layer of eye and is brownish black in colour. It is highly vascular and have pigments known as melanocytes which give colour to eye. Retina is the inner most and incomplete layer that extends up to ciliary body. The outer most layer of retina has two types of cells namely rods and cones.

Question 3.
Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre
(b) Depolarisation of the membrane of a nerve fibre
(c) Conduction of a nerve impulse along a nerve fibre
(d) Transmission of a nerve impulse across a chemical synapse
Answer:
(a) Polarization of the membrane of a nerve fibre:

• When a neuron is not conducting any impulse, i.e., at rest the axon membrane is more permeable to potassium ions (K⁺).
• It is impermeable to sodium ions (Na⁺), negatively charged proteins and Cl^– ions.
• Consequently, the axoplasm inside the axon contains a high concentration of K⁺ and negatively charged proteins and low concentration of Na⁺.
• In contrast, the fluid outside the axon contains a low concentration of K⁺, a high concentration of Na⁺ and thus forms a concentration gradient. These ionic gradients across the resting membrane are maintained by the active transport of ions by the sodium-potassium pump which transports 3 Na⁺ outwards for 2K⁺ into the cell.
• Asa result, the outer surface of the axonal ‘ membrane possesses a positive charge
  while its inner surface becomes negatively charged and therefore is polarised.
• The electrical potential difference across the resting plasma membrane is called resting potential.
• The electrical potential that occurs across the membrane of an axon, when stimulated by threshold stimulus is called depolarisation.

(b) Depolarisation of the membrane of a nerve fibre:

• When a stimulus is applied at a site on the polarised membrane, the membrane becomes more permeable to sodium ions (Na⁺) than to potassium ions (K⁺).
• The potential difference in the stimulated/ depolarised membrane is called an action potential.
• The action potential spreads like a wave along the membrane in the form of an impulse or spike.

(c)

Consider site A to be the site of excitation and hence depolarised. At sites immediately ahead, the axon (eg: site B) membrane has a .positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B. On the outer surface, current flows from site B to site A to complete the circuit of current flow. Hence, the polarity at the site is reversed, and an action potential is generated at site B. Thus, the impulse generated at site A arrives at site B. The sequence is repeated along the length of the axon and consequently, the impulse is conducted.

(d) A nerve impulse is transmitted from one neuron to another through junctions called synapses. At a chemical synapse, the membranes of the pre – and postsynaptic neurons are separated by a fluid-filled space called Synaptic cleft. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses. When an impulse arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.

The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane. This binding opens ion channels allowing the entry of ions which can generate a new potential in the post-synaptic neuron. The new potential developed may be either excitatory or inhibitory.

Question 4.
Draw labelled diagrams of the following:
(a) Neuron
(b) Brain
(c) Eye
(d) Ear
Answer:
(a) Neuron

(b) Brain
The human brain is well protected by the skull. Inside the skull, the brain is covered by cranial meninges consisting of three layers namely outer durometer, middle layer called arachnoid and inner layer pia mater. The brain is divided into three major parts.

• Forebrain
• Midbrain
• Hindbrain.

Forebrain:
The forebrain consists of cerebrum, thalamus and hypothalamus. Cerebrum is divided longitudinally into two halves, left and right cerebral hemispheres. The hemispheres are connected by a tract of nerve fibres called the corpus callosum. The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. It is referred to as grey matter due to greyish appearance. The cerebral cortex contains motor areas, sensory areas and association areas which are responsible for complex functions like intersensory associations, memory and communication.

Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. This layer is white in colour, hence called white matter. The cerebrum wraps around a structure called thalamus, which is a major co-ordinating centre for sensory and motor signalling. Hypothalamus lies at the base of the thalamus and contains several group of neurosecretory cells, which secrete hormones called hypothalamic hormones. The inner parts of cerebral hemispheres and a group of associated deep structures like the amygdala, hippocampus, etc. form a complex structure called the limbic lobe or limbic system.

(ii) Midbrain:
The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passes through the mid-brain. The dorsal portion of the midbrain consists of four round swellings (lobes) called Corpora quadrigemina. Midbrain and hindbrain form the brain stem.

(iii) Hindbrain:
Hindbrain comprises pons, cerebellum and medulla. Pons consists of fibre tracts that interconnect different regions of the brain. Cerebellum has very convoluted surface in order to provide additional space for many more neurons. The medulla of the brain is connected to the spinal cord.

(c) Eye:

The adult human eyeball is nearly a spherical structure. The wall of the eyeball is composed of three layers. The external layer called Sclera is composed of dense connective tissue. The middle layer choroid contains many blood vessels and looks bluish in colour. The choroid layer is thin over the posterior two-thirds of the eyeball, but it becomes thick in the anterior part to form the ciliary body. The ciliary body continues to form a pigmented and opaque structure called the iris.

The eyeball contains a transparent crystalline lens which is held in place by ligaments attached to the ciliary body. The aperture surrounded by the iris is called the pupil. The diameter of the pupil is regulated by the muscle fibre of iris. The inner layer is the retina and it contains three layers of cells called ganglion cells, bipolar cells and photoreceptor cells. Rods and cones are two types of photoreceptor cells. The optic nerves leave the eye and the retinal blood vessels enter it at a point medial to and slightly above the posterior pole of the eyeball. Photoreceptor cells are not present in that region and hence it is called the blind spot.

At the posterior pole of the eye lateral to the blind spot, there is a yellowish pigmented spot called macula lutea with a central pit called the fovea. The fovea is a thinned-out portion of the retina where only the cones are densely packed. It is the point where the visual activity is the greatest.

The space between the cornea and the lens is called the aqueous chamber and contains a thin watery fluid called aqueous humor. The space between the lens and the retina is called the vitreous chamber and is filled with a transparent gel called vitreous humor.

(d) Ear
Ear is divided into three major sections called the outer ear, the middle ear and the inner ear. The outer ear consists of the pinna and external auditory meatus. The external auditory meatus leads inwards and extends up to the tympanic membrane (eardrum). There are very fine hairs and wax- secreting sebaceous glands in the skin of the pinna and the meatus. The tympanic membrane is composed of connective tissues.

Covered with skin outside and with mucus membrane inside. The middle ear contains three ossicles called the malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane and the stapes are attached to the oval window of the cochlea. A Eustachian tube connects the middle ear cavity with the pharynx.

The fluid-filled inner ear called labyrinth consists of two parts, the bony and the membranous labyrinths. The bony labyrinth is a series of channels. Inside these channels lies the membranous labyrinth, which is surrounded by a fluid called perilymph. The membranous labyrinth is filled with a fluid called endolymph. The coiled portion of the labyrinth is called the cochlea.

The membranes consisting cochlea, the reissners and basilar, divide the surrounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibuli ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear.

The organ of Corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti. The basal end of the hair ceil is in close contact with the afferent nerve fibres. A large number of processes called stereocilia are projected from the apical part of each hair cell.

Above the rows of the hair cells is a thin elastic membrane called the tectorial membrane. The inner ear also contains a complex system called vestibular apparatus, located above the cochlea. The vestibular apparatus is composed of three semicircular canals and the otolith organ consisting of the saccule and utricle. Each semicircular canal lies in a different plane at right angles to each other. The membranous canals are suspended in the perilyrrph of the bony canals. The base of canals is swollen and is called ampulla, which contains a projecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a projecting ridge called macula.

Question 5.
Write short notes on the following:
(a) Neural coordination
(b) Forebrain
(c) Midbrain
(d) Hindbrain
(d) Retina
(f) Ear ossicles
(g) Cochlea
(h) Organ of Cortl
(I) Synapse
Answer:
(a) Neural co-ordination: Co-ordination is the process through which two or more organs interact and complement the functions of one another. In our body the neural system and the endocrine system jointly coordinate and integrate all the activities of the organs so that they function in a synchronized fashion. The neural system provides an organised network of point-to-point connections for quick co-ordinations. Neural – co-ordination includes mechanisms like transmission of nerve impulse, impulse conduction across a synapse and reflex action.

(b) Forebrain:
The forebrain consists of cerebrum, thalamus and hypothalamus. Cerebrum is divided longitudinally into two halves, left and right cerebral hemispheres. The hemispheres are connected by a tract of nerve fibres called corpus callosum. The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. It is referred to as grey matter due to greyish appearance. The cerebral cortex contains motor areas, sensory areas and association areas which are responsible for complex functions like intersensory associations, memory and communication.

Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. This layer is white in colour, hence called white matter. The cerebrum wraps around a structure called thalamus, which is a major co-ordinating centre for sensory and motor signalling. Hypothalamus lies at the base of the thalamus and contains several group of neurosecretory cells, which secrete hormones called hypothalamic hormones. The inner parts of cerebral hemispheres and a group of associated deep structures like amygdala, hippocampus, etc. form a complex structure called the limbic lobe or limbic system.

(c) Midbrain:
The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hind brain. A canal called the cerebral aqueduct passes through the mid-brain. The dorsal portion of the midbrain consists of four round swellings (lobes) called Corpora quadrigemina. Midbrain and hindbrain form the brain stem

(d) Hindbrain:
Hindbrain comprises pons, cerebellum and medulla. Pons consists of fibre tracts that interconnect different regions of the brain. Cerebellum has very convoluted surface in order to provide additional space for many more neurons. The medulla of the brain is connected to the spinal cord.

(e) Retina: Retina is the inner layer of the eye and contains three layers of cells; ganglion cells, bipolar cells and photoreceptor cells, (from inside to outside respectively). There are two types of photoreceptor cells, namely rods and cones. These cells contain light-sensitive proteins called photopigments. Daylight vision and colour vision are functions of cones and the twilight vision is the function of the rods. The rods contain a purplish-red protein called the rhodopsin or visual purple, which contains a derivative of Vitamin A. There are three types of cones that respond to red, green and blue lights.

(f) Ear ossicles: The ear ossicles seen in the middle ear are malleus, incus and stapes. They are attached to one another in a chain-like fashion. Malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.

(g) Cochlear: The coiled portion of the labyrinth is called cochlea. The membranes constituting cochlea, the Meissner’s and basilar, divide the surrounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibule ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear.

(h) Organ of Corti: The organ of Corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti. The basal end of the hair cell is in close contact with the afferent nerve fibres. A large number of processes called stereocilia are projected from the apical part of each hair cell. Above the rows of the hair cells is a thin elastic membrane called the tectorial membrane.

(i) Synapse: A nerve impulse is transmitted from one neuron to another through junctions called synapses. A synapse is formed by the membranes of a pre-synaptic neuron and a postsynaptic neuron, which may or may not be separated by a gap called the synaptic cleft. There are two types of synapses, namely, electrical synapses, the membranes of pre and post-synaptic neurons are in very close proximity. Electrical current can flow directly from one neuron into the other across these synapses.

The transmission of an impulse across electrical synapses is very similar to impulse conduction along a single axon. Impulse transmission across an electrical synapse is always faster than that across a chemical synapse. At a chemical synapse, the membranes of the pre and postsynaptic neurons are separated by a fluid-filled space called the synaptic cleft

Question 6.
Give a brief account of:
(a) Mechanism of synaptic transmission
(b) Mechanism of vision
(c) Mechanism of hearing
Answer:
(a) Mechanism of synaptic transmission: When an impulse reaches at the axon terminal, it stimulates the movement of the synaptic vesicles towards the synaptic cleft and releases their neurotransmitters. The neurotransmitter binds with the receptors found on the postsynaptic membrane. This binding opens ion channels allowing the entry of ions which can generate an action potential on the postsynaptic neuron.

(b) Mechanism of vision: The rays in visible wavelength focussed on the retina through the cornea and lens generate potentials (impulse) in rods and cones. The photo-sensitive compounds (photopigments) in the human eyes is composed of opsin and retinal. Light induces dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability to change.

As a result, potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells. These action potentials are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognised based on earlier memory and experience.

(c) Mechanism of hearing: The external ear receives sound waves and directs them to the eardrum. The eardrum vibrates in response to the sound waves and these vibrations are transmitted through the ear ossicles to the oval window. The vibrations are passed through the oval window on to the fluid of the cochlea, where they generate waves in the lymph.

The waves in the lymphs induce a ripple in the basilar membrane. These movements of the basilar membrane bend the hair cells, pressing them against the tectorial membrane. As a result, nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

Question 7.
Answer briefly:
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina.
Answer:
(a) Cones are responsible for colour vision.
They require brighter light than rods. There are three types of cones, maximally sensitive to long wavelength, medium- wavelength, and short-wavelength light (often referred to as red, green, and blue, respectively, though the sensitivity peaks are not actually at these colours). The colour seen is the combined effect of stimuli to, and responses from, these three types of cone cells. When these cones are stimulated equally, a sensation of white light is produced.

(b) The inner ear has three semi-circular canals forming cochlea. The inner ear also contains a complex system called vestibular apparatus, located above the cochlea.
The vestibular apparatus is composed of three semi-circular canals and the otolith organ consisting of the saccule and utricle. The base of canals is swollen and is called ampulla, which contains a proj ecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a proj ecting ridge called macula. The crista and macula are the specific receptors of the vestibular apparatus responsible for maintenance of balance of the body and posture.

(c) The light rays in visible wavelength focussed on the retina through the cornea and lens generate potentials (impulses) in rods and cones.
Light induces dissociation of the retinal from opsin resulting in changes in the structure of the opsin.
This causes membrane permeability changes. As a result, potential differences are generated in the photoreceptor cells.
These action potentials are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognised based on earlier memory and experience.

Question 8.
Explain the following:
(a) Role of Na⁺ in the generation of action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.
Answer:
(a) Role of Na⁺ in action potential:

• At the point of stimulation, the membrane becomes freely permeable to . sodium ions (Na⁺).
• As sodium ions enter the axoplasm, the interior becomes positively charged and the exterior is negatively charged. 9.
• The positive ions travel from the depolarised region to the next polarised region through the axoplasm and create an action potential there.
• It results in forming wave of excitation along with the nerve fibre which is called nerve impulse.

(b) Mechanism of generation of light-induced impulse in the retina:

• Light induces the dissociation of retinal (an aldehyde of vitamin A) and opsin (a protein), this results in change in the structure of opsin.
• The permeability of the membrane changes as a result of the above dissociation.
• The potential differences generated in the photoreceptor cells produce a signal that generates action potential in the bipolar neurons.
• These impulses/action potentials are transmitted by the optic nerve to the visual cortex.
• The neural impulses are analysed and image formed is recognised based on the earlier memory and experience.

(c) Mechanism through which a sound produces a nerve impulse in the inner ear :

• The sound waves vibrate the eardrum.
• The vibrations produced in response to these waves are transmitted through the ear ossicles to the oval window from where they reach the fluid of the cochlea.
• The waves produced in the perilymph and endolymph induce a ripple in the basilar membrane.
• The movements of the basilar membrane bend the hair cells which press them against the tectorial membrane.
• Nerve impulses are generated in the associated afferent neurons.
• The afferent fibres transmit the impulses via auditory nerves to the auditory cortex of the brain where the impulses are analysed and the sound is recognised.

Question 9.
Differentiate between:
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum
Answer:
(a) The myelinated nerve fibres are enveloped with Schwann cells, which form a myelin sheath around the axon. The gaps between two adjacent myelin sheaths are called nodes of Ranvier. Myelinated nerve fibres are found in spinal and cranial nerves.

Unmyelinated nerve fibre is enclosed by a Schwann cell that does not form a myelin sheath around the axon, and is commonly found in autonomous and somatic nervous systems.

(b) Dendrites are short fibres which branch repeatedly and project out of the cell body. It also contains Nissl’s granules. These fibres transmit impulses towards the cell body. The axon is a long fibre, the distal and which is branched. Each branch terminates as a bulb like structure called synaptic knob which possesses synaptic vesicles containing chemicals called neurotransmitters. The axon transmit nerve impulses away from the cell body to a synapse orto a neuromuscular junction.

(c)

Rods	Cones
(i) Responsible for twilight vision.	(i) Responsible for daylight and colour vision.
(ii) More in number	(ii) Less in number
(iii) Respond to lower light intensity.	(iii) Sensitive to bright light
(iv) Absent in fovea centralis	(iv) Present in the fovea
(v) Contain pigment Rhodopsin	(v) Contain pigment iodopsin
(v) Contain pigment	(v) Contain pigment

(d) The cerebrum wraps around a structure called the thalamus, which is major coordinating centre for sensory and motor signaling. It functions as a relay station. Hypothalamus lies at the base of the thalamus. This portion contains a number of centres which control body temperature, urge for eating and drinking. It also contains several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones.

(e)

Cerebrum	Cerebellum
(i)  Portion of fore brain. (ii) Seat of highest mental faculties, governs reasoning, learning, memory, intelligence (iii) Responds to heat, cold, pain, touch, light and pressure.	(i)  Portion of hind brain (ii) Regulates and coordinates the contraction of skeletal muscles. (iii) Maintains equili­brium and controls posture.

Question 10.
Answer the following:
(a) Which part of the ear determines the pitch of a sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock?
Answer:
(a) Organ of Corti
(b) Cerebrum
(c) Pineal gland

Question 11.
The region of the vertebrate eye, where the optic nerve passes out of the retina, is called the ………………….
(a) fovea
(b) Iris
(c) blind spot
(d) optic chiasma
Answer:
(c) blind spot

Question 12.
Distinguish between:
(a) afferent neurons and efferent neurons
(b) impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre
(c) aqueous humor and vitreous humor
(d) blind spot and yellow spot
(f) cranial nerves and spinal nerves.
Answer:
(a) The neuron which connects sense organs aw’ brain is called afferent neuron. The neuron which connects brain and effector organs or the concerned peripheral tissues  organs is called efferent neuron.

(b) In a myelinated neuron, impulse is transmitted by chemical method. It allows fast, saltatory movement of action potentials from node to node. In an unmyelinated neuron, impulse is transmitted by clerical mechanism. The conduction velocity of myelinated neurons vary linearly with axon diameter, whereas the speed of unmyelinated neurons vary roughly as the square root of the diameter of axons. Myelinated axon conduction is fast and energy efficient.

(c) The space between the cornea and the lens is called the aqueous chamber and contains a thin watery fluid called aqueous humor. The space between the lens and the retina is called the vitreous chamber and is filled with a transparent gel called vitreous humor.

(d) The optic nerves leave the eye and the retinal blood vessels enter it at a point medial to and slightly above the posterior pole of the eyeball called the blind spot. It doesn’t contain any photoreceptor cells (rods and cones). At the posterior pole of the eye lateral to the blind spot, is a yellowish pigmented spot called macula lutea (yellow spot). It contains cones only and the point where the visual acuity is the greatest

(e) Cranial nerves are the nerves that emerge directly from the brain stem in contrast to spinal nerves which emerge from segments of the spinal cord.

1st PUC Biology Neural Control and Coordination Additional Questions and Answers

1st PUC Biology Neural Control and Coordination One Mark Questions

Question 1.
What are meninges? (July 2011)
Answer:
The connective tissue membranes around brain are called meninges.

Question 2.
What is CNS?
Answer:
CNS or central nervous system is that includes brain and spinal cord.

Question 3.
What is Durometer?
Answer:
The outermost layer of meninges of CNS is Durometer.

Question 4.
What is corpus callosum?
Answer:
A transverse band of myelinated nerve fibers that connects cerebral hemispheres is called corpus callosum.

Question 5.
What is reflex arc? (June 2009)
Answer:
The path through which the electrochemical impulses are generated in response to a stimulus is called reflex arc.

Question 6.
What is reflex action?
Answer:
The quick, spontaneous and involuntary action induced by the nervous system in
response to a stimulus is called reflex action.

Question 7.
Which part of brain maintains body equilibrium? (Oct. 2000)
Answer:
Cerebellum is the centre for muscular coordination and equilibrium.

Question 8.
Name the protective covering of the brain. (April 2001)
Answer:
Meninges

Question 9.
Name the deep bridge of rieVve fibres which Joins cerebral hemispheres*. (Oct. 2003)
Answer:
Corpus callosum.

Question 10.
Name the structural and functional unit of nervous system. (July 2006)
Answer:
neuron

Question 11.
What is meant by co-ordination in our body?
Answer:
Co-ordination is the process through which two or more organs interact and complement the functions of one another.

Question 12.
What are afferent nerve fibers?
Answer:
The nerve fibres that transmit impulses from tissues / organs to the CNS are called afferent nerve fibres.

Question 13.
What are efferent fibres ?
Answer:
The nerve fibres that transmit regulatory impulses from the CNS to the concerned peripheral tissues / organs are called efferent nerve fibres.

Question 14.
Name the classification of autonomic neural system.
Answer:

• Sympathetic neural system
• Parasympathetic neural system

Question 15.
What are the major parts of a neuron ?
Answer:
Neuron is composed of cell body, dendrites and axon.

Question 16.
What are Nisei’s granules?
Answer:
Mr.’Nissl’s granules are certain granular bod¬ies present in the cytoplasm of the cell body of a neuron.

Question 17.
What are dendrites?
Answer:
Short fibres that branch repeatedly and project out of the cell body and also contain Nissl’s granules are called dendrites.

Question 18.
Where are neuro transmitters found ?
Answer:
Neurotransmitters are chemicals found in synaptic vesicles present in synaptic knob.

Question 19.
Name the three types of neurons.
Answer:
Multipolar, bipolar and unipolar neurons.

Question 20.
Where are multipolar neurons found in human body?
Answer:
Cerebral cortex …

Question 21.
What are nodes of Ranvier ?
Answer:
The gaps between two adjacent myelin sheaths of axon are called nodes of Ranvier.

Question 22.
Where exactly are synaptic vesicles located? What is their role ? (Foreign 2006)
Answer:
Synaptic vesicles are found in synaptic knob. They contain chemicals called neurotransmitters which are involved in transmission of impulses.

Question 23.
What are the charges on outer surface and Inner surface of axonal membrane during rest?
Answer:
Outer surface is positively charged and inner surface is negatively charged.

Question 24.
What are the functions of association areas?
Answer:
Association areas are responsible for complex functions like inter sensory associations memory and communication.

Question 25.
Write one function of limbic system in human brain. (Foreign 2006)
Answer:
Sexual behaviour, excitement, pleasure, rage, fear and motivation.

Question 26.
What is the canal passing through mid brain called?
Answer:
Cerebral aqueduct.

Question 27.
What is corpora quadrlgemina ?
Answer:
Corpora quadrigemina are four round swellings (lobes) found in the dorsal portion of midbrain.

Question 28.
Name the three regions of hindbrain.
Answer:
Pons, cerebellum and medulla.

Question 29.
What is the function of medulla oblongata.
Answer:
It contains centres which control respiration, cardiovascular reflexes and gastric secretions.

Question 30.
Which is the visible coloured portion of the eye ?
Answer:
Iris.

Question 31.
Name the external layer of the eye.
Answer:
Sclera

Question 32.
Name the area of retina which contains only cones and no rods. (Delhi 1997)
Answer:
Fovea

Question 33.
Why is blind spot devoid of the ability for vision?
Answer:
Blind spot does not contain any photoreceptor cells (rods and cones), hence devoid of vision.

Question 34.
Name the photosensitive compounds in human eye.
Answer:
Mropsin and retinal.

Question 35.
Name the fluid in which the membranous labyrinth of the inner ear floats. (Delhi 1997)
Answer:
Perilymph

Question 36.
Give the technical names of the auditory ossicles in their natural sequence. (All India 1997)
Answer:
Malleus, incus and stapes.

Question 37.
What connects the middle ear cavity with the pharynx?
Answer:
Eustachian tube.

Question 38.
Name the space within cochlea filled with endolymph.
Answer:
Scala media.

Question 39.
What does vestibular apparatus composed of?
Answer:
Three semicircular canals and otolith organ.

Question 40.
What is macula?
Answer:
The saccule and utricle contain a projecting ridge called macula.

Question 41.
Name the receptors responsible for body balance and posture.
Answer:
Crista and macula.

Question 42.
What is the function of Eustachian tube?
Answer:
It helps in equalising the pressures on either sides of the ear drum.

1st PUC Biology Neural Control and Coordination Two Marks Questions

Question 1.
What is reflex arc and reflex action.
Answer:
The path through which the electro-chemical impulses are generated in response to a stimulus is called reflex arc. It is the simplest functional unit of the nervous system by which an impulse produces a reflex action.The quick, spontaneous and involuntary action induced by the nervous system in response to a stimulus is called reflex action.

Question 2.
What are the functions of CSF.
Answer:

• It acts as a buffer, protecting the brain and spinal cord.
• It conveys nourishment to the tissues of the brain and spinal cord.
• It protects both brain and spinal cord from shocks and maintains constant pressure with in cranium.

Question 3.
Write the functions of hypothalamus.
Answer:

• Thermo regulation
• Biological clock system
• Autonomic nervous system control
• Sleep
• Controlling body temperature, appetite, secretion of pituitary.

Question 4.
What are Meninges? Mention different layers of Meninges.
Answer:
The connective tissue membranes around brain are called meninges. They protect the delicate nerve structure, carry the blood vessels to it and by the secretion of a fluid
(C S F) minimize any blow or concussion. There are three layers of Meninges;

• Durometer
• The arachnoid
• Piamater

Question 5.
What happens if the cerebellum is damaged. (Oct. 97)
Answer:
As the cerebellum is concerned with the maintenance of equilibrium of the body and coordination of contractions and relaxations of muscles, a damage to the, cerebellum leads to loss of the ability in muscular co-ordination and equilibrium in a person.

Question 6.
List any four functions of hind brain. (July 06)
Answer:
The functions of the hind brain are

• It regulates posture and postural activities.
• It plays an important part in muscular co ordination and maintenance of balance
• Cerebellar hemisphere controls muscle tone and posture on its side
• It regulates smooth and precise goal oriented movements.

Question 7.
Briefly give an account of somatic and autonomic nervous system.
Answer:
The PNS is divided into two divisions:

• Somatic neural system
• Autonomic neural system (ANS)

Somatic neural system relays impulses from the CNS to skeletal muscles while the autonomic neural system transmits impulses from the CNS to the involuntary organs and smooth muscles of the body. The ANS is further classified into sympathetic neural system and parasympathetic neural system.

Question 8.
What are the functions of hypothalamus?
Answer:

• It controls body temperature, urge for eating and drinking.
• It contains cells that secrete hormones called hypothalamic hormones.
• It is involved in the regulation of sexual behaviour, expression, emotional reactions and motivation.

Question 9.
Write the difference between cerebrum and cerebellum
Answer:

Cerebrum	Cerebellum
(a) It is a part of forebrain (b)  It is meant for memory, intelli­gence and control of voluntary move­ments	(a)  It is a part of hindbrain (b)  It controls the move­ments and help in maintaining body posture.

Question 10.
Why are gray matter and white matter contained In human nervous system name so? (Delhi 2006)
Answer:
Grey matter:
It contains neural cell bodies (spindle, pyramidal and stellate neurons) which give them greyish appearance.

White matter:
This region contains millions of axons with myelin sheath which gives them an opaque white appearance.

Question 11.
Name and differentiate between two types of synapses. (Foreign 2004)
Answer:
The two types of synapses present are:

• Electrical synapse
• Chemical synapse

Electical synapse	Chemical synapse
(a)  Electrical current flows directly from one neuron to other. (b) Synaptic cleft is very narrow (c) Impulse transmis­sion is faster (d) These are less com­mon	(a)  Signal transmissions involve chemical called neurotransmit­ters. (b) Synaptic cleft is wider. (c)  Impulse transmission is slower. (d) These are more com­mon.

Question 12.
Name the ear ossicles in the order of arrangement in humans. What role do they play in hearing?
Answer:
Malleus, Incus, and Stapes are the ear ossicles. The vibrations due to sound waves are transmitted through the ear ossicles to the oval window which are further transmitted to endolymph.

Question 13.
What is blind spot? Why is it so named? (Foreign 2003)
Answer:
The point where optic nerves leave the eye and the retinal blood vessels enter which is slightly above posterior pole of the eye ball is called blind spot. It lacks photoreceptor cells (rods and cones) and is devoid of vision. Hence called blind spot.

Question 14.
What happens when the membrane of a nerve cell carries out a sodium pump? (Delhi 2001)
Answer:
When the membrane of a nerve cell carries out a sodium pump, it transports 3 Na⁺ ions outwards for 2 K⁺ ions into the cell. As a result, the outer surface of the axonal membrane posses a positive charge while its inner surface is negatively charged and therefore polarised.

Question 15.
What are the events that take place at the point of simulation of an axon? (Delhi 1997)
Answer:
When a stimulus is applied at a site on the polarised membrane, it becomes freely permeable to Na⁺. This leads to a rapid influx of Na⁺ which causes the outer surface of the membrane to be negatively charged and the inner side becomes positively charged i.e., depolarisation takes place.

Question 16.
What constitutes the outer ear. Mention one function of each.
Answer:
The outer ear consists of the pinna and external auditory meatus.

• Pinna collects the vibrations in the air which produce sound.
• There are very fine hairs and wax secreting sebaceous glands in the skin of pinna and the meatus.

1st PUC Biology Neural Control and Coordination Three Marks Questions

Question 1.
Draw a diagram to show the path followed by the nerve impulse from the receptor to the effector in a spinal reflex arc. Label any six parts. (All India 2004 )
Answer:

Question 2.
Where are synaptic vesicles found? Name their chemical contents. What Is the function of these contents?
Answer:
Synaptic vesicles are found in a bulb like structure called synaptic knob. Synaptic vesicles contain chemicals called neurotransmitters. Neurotransmitters are involved in the transmission of impulses at the synapses. They bind to the receptors present on the post – synaptic membrane and generate new potential.

Question 3.
Explain briefly the structure and function of human middle ear. (All India 1997 C)
Answer:
The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain like fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. An Eustachian tube connects the middle ear cavity with the pharynx.

Functions:

• The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.
• Eustrachian tube helps in equalising the pressures on either sides of the ear drum.

Question 4.
Name the three types of neurons based on the number of axon and dendrites and give the location of their presence.
Answer:
(i) Unipolar neurons:

• Cell body with one axon.
• found in embryonic stage

(ii) Bipolar neurons:

• Cell body with one axon and one dendrite
• found in retina of the eye

(iii) Multipolar neurons:

• Cell body with one axon and two or more dendrites
• Found in cerebral cortex

Question 5.
Draw a diagram showing axon terminal and synapse and label any six parts
Answer:

Question 6.
Draw a neat labelled diagram of sectional view of cochlea.
Answer:

1st PUC Biology Neural Control and Coordination Five Marks Questions

Question 1.
List out the functions of different parts of the human brain. (April 97)
Answer:
The human brain has 3 main divisions namely Forebrain, Midbrain and Hindbrain. The Forebrain is composed (or subdivided into) of the cerebrum and diencephalon, the Midbrain is also called the mesencephalon with only one subdivision and the hind brain is composed of two subdivisions namely the cerebellem and medulla oblongata.
The functions of all these parts forming the brain are listed below:

(1) Forebrain (prosencephalon)

• Cerebrum – Is a functional center for intelligence, consciousness, voluntary control, memory.
•  Diencephalon- (Thalamus & hypothalamus) – While the thalamus acts as a main relay centre for conducting information, the hypothalamus controls the temperature of the body, appetite, fat metabolism, secretions of pituitary, sleep and the emotional states namely fear and anger.

(2) Midbrain (Mesencephalon) is the control centre for visual and auditory reflexes like adjustment of the ear to sound, pupil reflex, blinking etc.
(3) Hindbrain (Rhombencephalon)

• Cerebellum is the control centre for muscular coordination and equilibrium.
• Pons connects various parts of the brain with one another and contains the respiratory centres
• Medulla oblongata: contains the centres that control the heart beat, blood pressure, respiration and also centres that control swallowing, coughing and vomiting.

Question 2.
(i) A person unconsciously withdraws his hand suddenly with a jerk after touching hot plata- Draw a schematic diagram of the nervous pathway involved in the response.
(ii) What is such a response called?
Answer:
(i)

(ii) Such a response is called ‘reflex action’

Question 3.
Match the following:
I – II
(a) Depolarisation (i) Mid brain
(b) Polarisation (ii) Hind brain
(c) Corpus callosum (iii) Fore brain
(d) Corpora quadrigemina (iv) Action potential
(e) Pons (v) Resting potential
Answer:
(a) → (iv)
(b) → (v)
(c) → (iii)
(d) → (i)
(e) → (ii)

Question 4.
Fill In the blanks.
(a) The………….. fibres transmit impulses from tissues to the CNS and …………..fibres transmit Impulses from CNS to organs.
(b) The gaps between two adjacent myelin sheaths is called …………….
(c) …………….. are responsible for functions like memory and communication.
(d) The coiled portion of the labyrinth is called …………..
(e) ………….. contains hair cells that act as auditory receptors.
Answer:
(a) Afferent, Efferent
(b) Nodes of Ranvier
(c) Association areas
(d) Cochlea
(e) Organ of Corti

Question 5.
State whether the following statements are true / false. Correct the statements If false.
(a) Impulse transmission across electrical synapse is faster than that of chemical
synapse.
(b) During polarisation of axonal membrane, outer surface is negatively charged and inner surface is positively charged.
(c) Middle layer of cranial meninges is called pla mater.
(d) Cerebellum is responsible for controlling respiration and gastric secretions.
(e) Visible coloured portion of the eye Is Iris.
Answer:
(a) True.
(b) False; outer surface is positively charged and inner surface is negatively charged.
(c) False; middle layer is called arachnoid
(d) False; Medulla is responsible for controlling respiration and gastric secretions.
(e) True.

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1st PUC Biology Locomotion and Movement NCERT Text Book Questions and Answers

Question 1.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:

Question 2.
Define the sliding filament theory of muscle contraction.
Answer:
Sliding filament theory states that contraction of a muscle fibre takes place by the sliding of the thin filaments over thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Answer:
Steps in muscle contraction:
A nerve impulse arriving at the neuromuscular junctions initiates the contractile response.

• A neurotransmitter released at the neuromuscular junction enters the sarcomere through its membrane channel.
• The opening of the channel also results in the inflow of Na+ ions inside the sarcomere and generates an action potential that travels along the entire length of the muscle fibres.
• The sarcomere reticulum releases Ca++ ions, which bind with the specific sites present on the troponin component of the thin filament.
• Asa result of conformational changes in the troponin, the active sites on the F-actin are exposed.
• These are the active sites specific to the myosin head, which exhibits myosin-dependent ATPase activity.
• The myosin heads act as hooks and attach to F-actin to form cross-bridges.
• When the muscle is stimulated to contract, the cross-bridges move, pulling the two filaments past each other.
• When thousands of actin and myosin filaments interact this way the entire muscle cell shortens.

This concept is the sliding filament theory.

Question 4.
Write true or false. If false change the statement so that it is true.

1. Actin is present in the thin filament
2. H-zone of striated muscle fiber represents both thick and thin filaments.
3. The human skeleton has 206 bones.
4. There are 11 pairs of ribs in man.
5. The sternum is present on the ventral side of the body.

Answer:

1. True
2. False; H – zone represents thick filaments.
3. True
4. False; There are 12 pairs of ribs.
5. True.

Question 5.
Write the difference between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle
Answer:
(a) Actin filaments are thinner as compared to the myosin filaments. Actin filament exists in two forms namely globular actin or G actin and fibrous actin of F- actin. It also contains two proteins namely tropomyosin and troponin. Myosin filament or myofibril consists of two portions, head, and tail. The head is formed of heavy meromyosin (HMM) and the tail is formed of light meromyosin (LMM). The myosin head possesses contractile property as well as ATPase activity.

(b) Red and White muscles
Answer:

Red Muscle	White Muscle
(a) Red in colour due to the presence of large amount of myoglobin (b) Thin (c) Slow and sustained contraction (d) Depend on aerobic metabolism for energy (e) Presence of large amount  of mitochondria (f) Perform sustained work for a long time. (g) Blood capillaries more, eg: Extensor muscle on the back of the human body.	(a) Lesser amount of myoglobin, hence white in colour (b) Thick (c) Contraction rate is fast (d) Depend  on anaerobic metabolism for energy (e) Presence of small amount of mitochondria (f) Perform strenuous work for a short time. (g) Blood capillaries less, eg: Muscles of the eyeball.

(c) Pectoral and Pelvic girdle
Answer:

Pectoral girdle	Pelvic girdle
(i) Connects forelimb with axial skeleton	(i) Connects leg with the axial skeleton
(ii) Each half is formed of two bones, clavide and scapula	(ii) Each half is formed  of three bones ilium, ischium and pubis
(iii) Two halves are not fused	(iii) Two halves are fused
(iv) Found in backside of thorax	(iv) Found in hip region
(v) Cavity is known as glenoid cavity.	(v) A cavity is known as acetabulum

Question 6.
Match Column I with Column II:

Column I – Column – II

(a) Smooth muscle – (I) Myoglobln
(b) Tropomyosin – (II) Thin filament
(c) Red muscle – (III) Sutures
(d) Skull – (lV) Involuntary
Answer:
(a) (iv)
(b) (ii)
(c) (i)
(d) (iii)

Question 7.
What are the different types of movements exhibited by the cells of the human body?
Answer:
Cells of the human body exhibit three main types of movements, namely, amoeboid, ciliary and muscular. Some specialized cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is affected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in the amoeboid movement.

Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement. Movement of our limbs, jaws, tongue, etc, requires muscular movement. The contractile property of muscles is effectively used for locomotion and other movements by human beings and the majority of multicellular organisms. Locomotion requires a perfect coordinated activity of muscular, skeletal, and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Answer:
Skeletal muscle: Skeletal muscles are closely associated with skeletal components of the body. They have a striped appearance under the microscope and hence are called striated muscles. As their activities are under the voluntary control of the nervous system, they are known as voluntary muscles too. They are primarily involved in locomotory actions and changes in body postures.

Cardiac muscles: Cardiac muscles are the muscles of the heart. Many cardiac muscle cells assemble in a branching pattern to form a cardiac muscle. Based on appearance, cardiac muscles are striated. They are involuntary in nature as the nervous system does not control their activities directly.

Question 9.
Name the type of Joint between the following:

1. Atlas/axis
2. Carpal/metacarpal of the thumb
3. Between phalanges
4. Femur/acetabulum
5. Between cranial bones
6. Between pubic bones in the pelvic girdle

Answer:

1. Pivot joint
2. Saddle joint
3. Gliding joint
4. Ball and socket joint
5. Fibrous joint
6. Cartilaginous joint

Question 10.
Fill in the blank spaces:

1. All mammals (except a few) have ……………… cervical vertebra.
2. The number of phalanges in each limb of humans………………
3. A thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ……………… and ………………
4. In a muscle fibre Ca++ is stored in ………………
5. ……………… and pairs of ribs are called floating ribs……………….
6. The human cranium Is made of ………………bones.

Answer:

1. 7
2. 14
3. Troponin and Tropomyosin
4. Sarcoplasmic reticulum
5. 11 and 12
6. 8

1st PUC Biology Locomotion and Movement Additional Questions and Answers

1st PUC Biology Locomotion and Movement One Mark Questions

Question 1.
Name two sites in our body where ciliary movement is seen.
Answer:
Fallopian tube and trachea.

Question 2.
Give two examples of movements that involve muscular movement.
Answer:
Movement of limbs, jaw, tongue etc.

Question 3.
Why are striated muscles called skeletal muscles?
Answer:
Striated muscles are attached to bones and so-called skeletal muscles.

Question 4.
Mention two sites on our body where striated muscles are present.
Answer:
Limbs and tongue.

Question 5.
Why are smooth muscles called visceral muscles?
Answer:
Smooth muscles are located in the inner walls of hollow visceral organs of the body like the alimentary canal, reproductive tract etc. So they are known as visceral muscles.

Question 6.
Name the connective tissue layer of muscle bundles.
Answer:
Fascia.

Question 7.
Where is ciliary movement mostly seen in our body?
Answer:
Ciliary movement occurs in most of our internal tubular organ which are lined by ciliated epithelium.

Question 8.
Name the three types of muscles in humans.
Answer:
Skeletal, visceral and cardiac.

Question 9.
Name the cells which exhibit amoeboid movement.
Answer:
Macrophages and leucocytes in blood.

Question 10.
What are myofibrils?
Answer:
Large number of parallelly arranged filaments in the sarcoplasm are called myofibrils.

Question 11.
What is the striated appearance of myofibrils due to?
Answer:
Striated appearance of myofibril is due to the distribution pattern of two proteins, actin and myosin.

Question 12.
Name the functional Unit of muscles.
Answer:
Sacromere.

Question 13.
How are action and myosin filaments arranged during the resting state of muscle?
Answer:
In a resting state, the edges of the thin filaments on either side of the thick filaments, partially overlap the free ends of the thick filaments.

Question 14.
What is a H – zone?
Answer:
Central part of the A- band not overlapped by the thin filaments is called H – zone.

Question 15.
Name the protein which is distributed at regular intervals on the tropomyosin.
Answer:
Troponin.

Question 16.
What is the function of troponin?
Answer:
In the resting state a subunit of troponin masks the active binding sites for myosin on the actin filaments

Question 17.
Name the monomers of myosin filament.
Answer:
Meromyosins.

Question 18.
What are the two parts of meromyosins?
Answer:
Meromyosin contains two parts:

• Globular head with a short arm called heavy meromyosin (HMM)
• A tail called Light meromyosin (LMM)

Question 19.
What is a cross arm?
Answer:
The head of meromyosin projects outwards at regular distance and angle from each other from the surface of polymerised myosin filament which is known as cross arm.

Question 20.
What causes fatigue in muscles?
Answer:
Repeated activation of the muscles lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen causing fatigue.

Question 21.
What Is myoglobin?
Answer:
Myoglobin is a red coloured oxygen storing pigment present in muscles.

Question 22.
Name two specialised connective tissues.
Answer:
Bone and cartilage

Question 23.
How many bones does a human being contain ?
Answer:
206 bones.

Question 24.
What salts are present In cartilage?
Answer:
Chondroitin salts.

Question 25.
How many bones does axial skeleton comprise of?
Answer:
80 bones.

Question 26.
What constitute axial skeleton?
Answer:
Skull, vertebral column, sternum and ribs.

Question 27.
Name the two set of bones in the skull.
Answer:
Cranial bones and facial bones.

Question 28.
What are ear ossicles made of?
Answer:
Malleus, Incus and Stapes.

Question 29.
How does skull region articulate with the superior region of the vertebral column?
Answer:
With the help of occipital condyles.

Question 30.
What is vertebral column formed of?
Answer:
Vertebral column is formed by 26 serially arranged units called vertebrae.

Question 31.
What is sternum?
Answer:
Sternum is a flat bone on the ventral midline of thorax.

Question 32.
What is a hyoid?
Answer:
Hyoid is a U- shaped bone present at the base of the buccal cavity.

Question 33.
Why are ribs called bicephalic?
Answer:
Ribs contain two articulation surface on its dorsal end and hence called bicephalic.

Question 34.
Name the major components of appendicular skeleton.
Answer:
The bones of the limbs and their girdles.

Question 35.
Name the longest bone in human body. (All India 2005)
Answer:
Femur (thigh bone)

Question 36.
What is acromion?
Answer:
The dorsal, flat, triangular body of scapula has a spine which projects as a flat, expanded process called the acromion.

Question 37.
What bones help In the articulation of the upper and the lower limbs with the axial skeleton.
Answer:
Pectoral and Pelvic girdle respectively.

Question 38.
What is the technical term for collar bone?
Answer:
Clavicle.

Question 39.
Name the cavity in the pectoral girdle with which head of humerus articulates.
Answer:
Glenoid cavity.

Question 40.
What Is meant by coxal bones?
Answer:
The two halves of the pelvic girdle are called coxal bones.

Question 41.
What is acetabulum in the pelvic girdle?
Answer:
Acetabulum is a cavity at the point of fusion of coxal bones to which the thigh bone articulates.

Question 42.
Name the components of coxal bone.
Answer:
Illium, ischium and pubis.

Question 43.
What are the Joints?
Answer:
Joints are points of contact between bones, or between bones and cartilages.

Question 44.
Name the classification of joints.
Answer:
Fibrous, cartilaginous and synovial.

Question 45.
What are pubic symphysis?
Answer:
Public symphysis is the cartilaginous joint between the two halves of the pelvic girdle on the ventral side.

Question 46.
Which type of joint Is the knee joint?
Answer:
Hinge joint.

Question 47.
How are synovial joints characterised ?
Answer:
Synovial joints are characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones.

Question 48.
What type of Joint is present between carpal and metacarpal of thumb ?
Answer:
Saddle joint.

Question 49.
What causes gouty arthritis In human ? (All India 2005)
Answer:
lnflammation of joints due to accumulation of uric acid crystals.

1st PUC Biology Locomotion and Movement Two Marks Questions

Question 1.
Mention four special properties of muscles.
Answer:

• Contractility
• Excitability
• Extensibility
• Elasticity

Question 2.
Give two examples of ciliary movement in our body.
Answer:

• The coordinated movements of cilia in the trachea help in removal of dust particles and foreign substances inhaled along with atmospheric air.
• Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

Question 3.
Differentiate between striated and smooth muscles.
Answer:

Striated	Smooth
(i) These are found attached to the skeletal elements	(i) They are found in the wall of visceral organs.
(ii) Striations are prominent	(ii) Striations are absent
(iii) They are voluntary in functions.	(iii) They are involuntary in function
(iv) Cells are multinucleate	(iv) Cells are uninucleate
(v) They are innervated by voluntary nervous system.	(v) They are innervated by autonomic nervous system

Question 4.
Bring out the differences between cardiac muscles and smooth muscles.
Answer:

Cardiac	Smooth
(i)  They are found only in the wall of the heart. (ii)   Striations are faintly seen (iii)  Cells are cyclindrical and branched (iv)  Intercalated discs are present	(i) They are found in the wall of visceral organs. (ii) Striations are absent (iii) Cells are spindle shaped (iv)  Intercalated discs are absent

Question 5.
What is sarcoplasmic reticulum? What is its function?
Answer:
The endoplasmic reticulum of striated muscle fibres is called sarcoplasmic reticulum. It is the storehouse of calcium ions.

Question 6.
What are A-band and I-band?
Answer:
Each myofibril has alternate dark and light bands on it. The striated appearance of myofibril is due to the distribution pattern of two important proteins – Actin and myosin. The light bands contain actin and is called I-band on isotropic band, whereas the dark band called ‘A1 band or Anisotropic band contains myosin.

Question 7.
Differentiate between bone and cartilage.
Answer:

Bone	Cartilage
(i) It is a hard or rigid connective tissue. (ii) The matrix is deposited with calcium salts. (iii) One osteocyte is found in a lacuna. (iv) Haversian system are present.	(i) It is a semi-rigid connective tissue (ii) The matrix contains chondroitin salts. (iii) 2 to 4 chondrocytes are present in lacuna. (iv) Haversian systems are absent.

Question 8.
What are the two principal divisions of human skeletal system ? How many bones are present in each of them and what do they constitute of ?
Answer:
(i) Axial skeleton

• 80 bones
• constitute of skull, vertebral column,
• sternum and ribs.

(ii) Appendicular skeleton

• 126 bones
• Constitute of bones of the limbs and their girdles.

Question 9.
What are the functions performed by vertebral column ?
Answer:

• It protects the spinal cord;
• It supports the head.
• It serves as the point of attachment for the ribs.
• Provides place for attachment of muscles of the back

Question 10.
Describe the vertebra chondral ribs.
Answer:
The 8th, 9th and 10th pairs of ribs are called vertebra – chondral ribs.
They remain attached dorsally to the respective thoracic vertebrae and ventrally to the sternum through the seventh rib.

Question 11.
Name the bones of palm. How many of them are present in each palm?
Answer:
Bones of palm are called metacarpals. There are five of them in each palm.

Question 12.
Explain a synovial joint Give two examples.
Answer:
Synovial joints are characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones. Such an arrangement allows considerable movement. These joints help in locomotion and many other movements. Ex: Ball and socket joint, Pivot joint.

Question 13.
Why can red muscle fibres work for a prolonged period, while a white muscle fibre suffers from fatigue soon? (Delhi 2005 C)
Answer:
Red muscle fibres contain a pigment myoglobin which stores oxygen. These muscles also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production. Hence they can work for a prolonged period without getting fatigue.

Whereas, white muscle fibres possess very less quantity of myoglobin and very few mitochondria in them. Hence they depend on anaerobic process for energy and get fatigue soon.

Question 14.
What is the role of calcium ions In muscle contraction ? (All India 2004)
Answer:

• Calcium ions bind to specific sites on the troponin component of thin (actin) filaments.
• As a result, conformational change occurs in the troponin molecules and the active sites of F- actin are exposed. These are the sites Specific to myosin head that shows Mg dependent ATP- ase activity.
• When calcium ions are pumped back into the sarcoplasmic reticulum, the troponin component becomes free to mask the active sites for myosin head. This leads to breakage of cross bridges and the actin filaments slide out of A – band.

Question 15.
List any four functions of skeleton in higher animals. (Delhi 1999)
Answer:

• It plays a vital role in movement and locomotion.
• It provides protection to many vital organs.
• It series as reservoir of calcium and phosphates.
• The bone marrow of long bones produces blood cells.

1st PUC Biology Locomotion and Movement Three Marks Questions

Question 1.
Explain giving one example of each, the three types of joints In human skeletons based on the capacity of movement. (Delhi 2006)
Answer:
(i) Immovable / Fibrous joint
The articulating bones are very tightly held with the help of white fibrous connective tissue and there is no space between the bones. Eg: Surtures between skull bones.

(ii) Slightly movable / Cartilaginous joints: The bones involved are joined together with the help of cartilages and allows limited movement. Eg: Joint between adjacent vertebrae in the vertebral column

(iii) Freely movable/synovial joints:
These joints are characterised by the presence of fluid filled synovial cavity between the articulating surfaces of two bones which allows considerable movement. Eg: Ball and socket joint, pivot joint etc.

Question 2.
What are the three types of muscle tissue? Write two characteristic points about the structure of each of them?
Answer:
Three types of muscle tissue are:
(a) Striated muscles

• These are skeletal muscles and are closely associated with the skeletal components of the body.
• Their activities are under the voluntary control of the nervous system.
• They are involved in locomotory actions and changes of body postures.

(b) Visceral muscles:

• They are located in the inner walls of hollow visceral organs like alimentary canal, reproductive tract etc.
• They do not exhibit any striation and are smooth if appearance.
• Their activities are not under the voluntary control of the nervous system.

(c) Cardiac muscles:

• They are the muscles of the heart and are striated.
• Cardiac muscle cells assemble in a branching pattern to form a cardiac muscle.
• They are involuntary in nature.

Question 3.
How many vertebrae In all, do we have? Categorize them on the basis of their location giving the specific number in each category. (Foreign 2004)
Answer:
We have a total of 26 vertebrae in our body.

• Cervical vertebrae (7) in the neck region.
• Thoracic vertebrae (12) present in the thoracic region of the trunk.
• Lumbar vertebrae (5) present in the abdominal region.
• Sacrum (1) is the triangular bone at the end of vertebral column.
  Coccygeal (1) which is vestigial tail bone in human being.

Question 4.
Describe the structure of sarcomere.
Answer:
The portion of the myofibril between two successive ‘z’ lines is called as sarcomere. It is the functional unit of muscle contraction. A sarcomere consists of a dark band in the centre and light bands on either side. The dark band or Anisotropic band (A band) is made of myosin and the light band on Isotropic band (I- band) is made of actin filaments.

Along the centre of the I- band is an elastic fibre called z – line, that bisects the band. Their filaments are attached firmly to the z – line. The myosin filaments are also held together in the centre by a thin fibrous membrane called M – Line. The central part of A band, not over lapped by thin filaments (I- band) is called a H – zone.

Question 5.
Draw a labelled diagram of vertebral column showing different regions.
Answer:

Question 6.
Name the different bones of forelimb and their number in human beings.
Answer:

• Humerus (1)
• Ulna (1)
• Radius (1)
• Carpals (8)
• Metacarpals (5)
• Phalanges (14)

Question 7.
Draw a neat labelled diagram of forelimb indicating all the bones.
Answer:

Question 8.
Name the A different bones of hind limb and their number in human beings.
Answer:

• Femur (1)
• Tibia (1)
• Fibula (1)
• PateNa (1)
• Tarsals (7)
• Metatarsals (5)
• Phalanges (14)

Question 9.
Draw a neat labelled diagram of hind limb indicating ail the bones.
Answer:

1st PUC Biology Locomotion and Movement Five Marks Questions

Question 1.
Explain the sliding filament theory of the mechanism of muscle contraction.
Answer:

Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fiber takes place by the sliding of the thin filaments over the thick filaments.

‘Muscle Contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron along with the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor – end plate. A neural signal reaching this junction releases a neurotransmitter which generates an action potential in the sacrolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm.

Increase in Ca⁺⁺ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin filaments towards the centre of ‘A’ band. The Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e contraction. During contraction the T bands get reduced, whereas the ‘A’ bands retain the length.

The ATP is again hydrolysed by the myosin head and the cycle of cross bridge formation and breakage is repeated causing further sliding. The process continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic Cisternae resulting in the masking of actin filaments. This causes the return of Z lines back to their original position, i.e. relaxation.

Question 2.
Briefly give an account of any five disorders of muscular and skeletal system.
Answer:

• Myasthenia gravis: Auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscles.
• Muscular dystrophy: Progressive degeneration of skeletal muscle mostly due to genetic disorder.
• Tetany : Rapid spasms in muscle due to low calcium ions in body fluid.
• Gouty arthritis: Inflammation of joints due to accumulation of uric acid crystals.
• Osteo porosis: Age related disorder characterised by decreasedbdhMfiass and increased chances of fractures. Decreased levels of estrogen is a common cause.

Question 3.
Represent diagrammatically the stages in cross bridge formation, rotation of head and its
Answer:

Question 4.
Draw a neat diagram of human skull labelling any 10 parts.
Answer:

Question 5.
Name different type of synovial Joint giving one example of each type.
Answer:

• Ball and Socket joint. Eg: between humerus and pectoral girdle.
• Hinge joint. Eg: Knee joint and Elbow joint.
• Pivot joint. Eg: between atlas and axis
• Gliding joint. Eg: between the carpals
• Saddle joint Eg: between carpal and metacarpal of thumb.

Question 6.
Draw a neat labelled diagram of
(a) An actin filament
(b) Myosin
Answer:

You can Download Chapter 13 Hydrocarbons Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 13 Hydrocarbons

1st PUC Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
What are hydrocarbons ?
Answer:
Organic compounds containing only carbon and hydrogen are called hydrocarbons.

Question 2.
What are alkanes ?
Answer:
Alkanes are saturated aliphatic hydrocarbons containing carbon to carbon single bonds.

Question 3.
Give the general formula of alkanes.
Answer:
The general formula of alkanes is C_nH_2n+2

Question 4.
What are saturated hydrocarbons ?
Answer:
The hydrocarbons containing only carbon – carbon single bonds saturated hydrocarbons.

Question 5.
What is the order of reactivity of halogens towards alkanes.
Answer:
The reactivity of halogens towards alkanes varies as Flourine > Chlorine > Bromine > Iodine i.e., F > Cl > Br > I.

Question 6.
Why alkanes are called parafins ?
Answer:
Because of low reactivity.

Question 7.
What is the nature ofC-C bond in alkanes?
Answer:
These are sigma (σ) bonds.

Question 8.
What type of structural isomerism is shown by alkanes?
Answer:
Chain Isomerism.

Question 9.
What is Photochemical reaction ?
Answer:
Reaction carried in the presence of light.

Question 10.
Which class of organic compounds are potential carcinogen?
Answer:
Polynuclear aromatic hydrocarbons.

Question 11.
What is huckel Rule ?
Answer:
It states that a compound is said to be aromatic. If it has (4n +2) p electrons which are de-localised where n = 0,1, 2, 3, ………

Question 12.
Name the scientist who first isolated benzene.
Answer:
Faraday isolated benzene in 1825.

Question 14.
What is decarboxylation ?
Answer:
The process of removing carbon dioxide from sodium salts of acid with the help of soda lime.

Question 15.
What is electrophilic subsitution reaction ?
Answer:
Those reaction in which weaker electrophile is replaced by stronger electrophile are called electrophilic substitution reaction.

Question 16.
Why is addition reaction of bromine to benzene difficult?
Answer:
Due to delocalisation of π electrons. It does not have pure double bond.

Question 17.
What are arenas? Give general formula of Monocyclic arenas.
Answer:
Arenes are aromatic hydrocarbons. General formula of arenas is C_nH_2n-6

Question 18.
What are unsaturated hydrocarbons?
Answer:
The hydrocarbons which contain atleast one carbon to carbon double bond or triple bond are called unsaturated hydrocarbons.

Question 19.
What type of isomerism is shown by alkenes ?
Answer:
Chain isomerism.

Question 20.
What are alkenes ?
Answer:
Alkenes are unsaturated aliphatic hydrocarbons containing one carbon to carbon double bond.

Question 21.
Give the general formula of alkenes.
Answer:
The general formula of alkenes is C_nH_2n.

Question 21.
Why are alkenes and alkenes called unsaturated hydrocarbons?
Answer:
Due to presence of C = C & C ≡ C bonds respectively.

Question 22.
What is the state of hybridisation of carbon atoms in alkenes ?
Answer:
Sp²

Question 23.
A compound decolourise yellow colour of bromine. What does it show ?
Answer:
It is unsaturated in nature.

Question 24.
What is Lindlar’s catalyst ?
Answer:
Palladium supported over calcium carbonate and deactivated with quinoline is called Lindlar’s Catalyst.

Question 25.
Name the hydrocarbon which contains acidic hydrogen.
Answer:
Ethylene (HC ≡ CH).

Question 26.
Give one hydrocarbon which delcolourises alk KMnO₄.
Answer:
Ethene (CH₂ = CH₂)

Question 27.
Name the reagent which brings about dehydrohalogination.
Answer:
Alcoholic potash.

Question 28.
What is the major product formed when propene reacts with hydrogen bromide ?
Answer:
2 – bromopropane.

Question 29.
Name the product formed when HBr reacts with 2 – methyl propene.
Answer:

Question 30.
What are alkynes?
Answer:
Alkynes are unsaturated aliphatic hydrocarbons containing one carbon to carbon triple bond.

Question 31.
Give the general formula of alkynes.
Answer:
The general formula of alkynes is C_nH_2n-2

Question 32.
What is the hybridisation of carbon atoms in acetylene ?
Answer:
sp

Question 33.
What is a vicinal dihalide ?
Answer:
A compound with halogen atoms on adjacent carbon atom.

Question 34. Give the example of vicinal dihalide.
Answer:

Question 35.
Give an example of geminal dihalide.
Answer:
CH₃ – CHCl₂ (1,1 – dichlorethane)

Question 36.
Give an example of alkyne.
Answer:
Acetylene CH ≡ CH

Question 37.
Why are alkenes called olefins ?
Answer:
Because they are oily forming compounds i.e., most of the oils are unsaturated.

Question 38.
Give an addition reaction of alkynes.
Answer:
CH ≡ CH + HBr → CH₂ ≡ CH – Br (Vinyl Bromide)

Question 39.
Give the resonance structures of benzene.
Answer:

Question 40.
Name the electrophiles in the following reactions

1. Chlorination
2. Nitration
3. Sulphonation
4. Friedel Craft’s methylation

Answer:

1. Chlorination – Cl⁺
2. Nitration – \(\mathrm{NO}_{2}^{+}\)
3. Sulphonation – SO₃ .
4. Friedel Crafts methylation – \(\mathrm{CH}_{3}^{+}\)

Question 41.
Write the Geometrical isomers of CHBr = CHBr
Answer:

Question 42.
Which catalyst is most effective even in the polymerization of ethane to polythene ?
Answer:
Ziegler – Nata Catalyst

Question 43.
What is the compostion of Zeigler Natta Catalyst?
Answer:
Triethylaluminium and titanium tetracholride inert solvent.

Question 44.
Name one Carcinogenic compound.
Answer:
Anthracene of Benzanthracene.

Question 45.
What is the number of a and n bonds in N ≡ C-CH = CH- C ≡ N?
Answer:
There are 7 σ bonds and 5 π bonds.

Question 46.
Name the compound formed when heptane is subjected to aromatisation.
Answer:
Toulene.

Question 47.
What is the cause of Geometrical Isomerism ?
Answer:
The restricted rotation around carbon to carbon double bond

Question 48.
What is cracking of hydrocarbon ?
Answer:
Cracking is a process of breaking higher hydrocarbons by heating.

Question 49.
What is the trade name of Benzene Hexacarbon ?
Answer:
Gammaxane.

1st PUC Chemistry Hydrocarbons Two Marks Questions and Answers

Question 1.
What is the action of ethane with bromine in carbon tetra chloride?
Answer:
Ethene reacts with bromine in carbon tetrachloride to give 1, 2 – dibromethane. Orange colour of bromine is discharged.

Question 2.
What is the action of hydrogen halide on alkene?
Answer:
Alkene reacts with hydrogen halide to form alkyl halide. The addition is governed by Markownikoff’s rule.

Question 3.
How do you obtain an alkene from alkyne ?
Answer:
Alkynes react with hydrogen in the presence of (Palladium supported over calcium carbonate and deactivated with quinoline) Lindlar’s catalyst, to give alkene.

Question 4.
How do you convert ethene to ethane?
Answer:
When a mixture of ethene and hydrogen is passed over heated nickel catalyst at 200°C, ethane is obtained.
\(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{H}_{2} \frac{\mathrm{Ni}}{200^{\circ} \mathrm{C}} \mathrm{CH}_{3}-\mathrm{CH}_{3}\)

Question 5.
How is propene obtained from propyl chloride ?
Answer:
When propyl chloride is heated with alcoholic potash propene is obtained.
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Cl}+\mathrm{KOH}_{(\mathrm{alc})} \stackrel{\Delta}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\).

Question 6.
What is Wurtz reaction ? Give example.
Answer:
When alkyl halides are heated with sodium metal in ether medium higher alkanes are formed. This reaction is known as Wurtz reaction and employed for the synthesis of higher alkanes containing even number of carbon atoms.

Question 7.
How is Methane prepared from sodium acetate ?
Answer:
Decarboxylation : When sodium salt of carboxylic acid is heated with soda lime (mixture of sodium hydroxide and calcium oxide), an alkane containing one carbon atom less than parent carboxylic acid is formed. This reaction is called decarboxylation.

Question 8.
How is alkane prepared from Kolbe’s electrolytic method ?
Answer:
A concentrated sodium salt of a carboxylic acid is electrolysed. The alkane is formed at the anode as one of the products. This process is called Kolb’e electrolysis.

Question 9.
What is Pyrolysis ? Give example.
Answer:
The decomposition of higher alkane into a mixture of lower alkanes, alkenes, etc by the application of heat is called pyrolysis or cracking.

Question 10.
What is Aromatisation ? Give example.
Answer:
Hexane or higher akanes when heated in presence of vanadium pentoxide (V₂O₅), molybdenium oxide (MO₂O₃) or chromium oxide (Cr₂O₃) supported on alumina (Al₂O₃) at 800 K and 10-20 atm pressure give benzene or its alkyl derivatives with the liberation of hydrogen. This dehydrogenation process which involves cyclis^ion of alkanes is known as aromatization.

Question 11.
What is isomerisation ? Give example.
Answer:
When unbrached alkanes are heated with anhydrous aluminium chloride and hydrogen chloride, isomeric branched alkanes are formed. This process is called isomerisation.

Question 12.
What is Geometrical isomerism ?
Answer:
Compounds having same molecular formula and structural formula but differ in the spatial arrangement of atoms or group of atoms due to restricted rotation of carbon carbon double bond.

Question 13.
What are the conditions for Geometrical isomerism ?
Answer:

• The molecule should contain double bond.
• Each carbon atom joined by the double bond, must be attached to two different groups.

Question 14.
What is Ozonalysis ? Give example.
Answer:
The ozonolysis of alkenes involves addition of ozone molecule to carbon-carbon double bond of alkenes to form ozonide. This follows hydrolysis of the ozonoide by H₂O/Zn to yield corresponding aldehydes and / or ketones. The addition of ozone to alkene followed by hydrolysis is known as ozonolysis.

Question 15.
How is ethene prepared ?
Answer:
When 1, 2 – dihaloalkanes (vicinal dihalides) are heated with zinc metal in presence of methanol corresponding alkenes are formed. This reaction is called dehalogenation.
\(\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{Zn} \frac{\mathrm{CH}_{3} \mathrm{OH}}{\text { heat }} \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{ZnBr}_{2}\)

Question 16.
What are the uses of LDPE and HDPE ?
Answer:
LDPE is used for

• making transparent films for packaging garments and food,
• as an insultor.

HDPE is used for

• making crates and large boxes
• production of carry bags, containers, house hold articles and toys.

Question 17.
What happens when Ammonical solution of silver Nitrate treated with Acetylene ?
Answer:
Acetylene when passed through an ammonical solution of silver nitrate (Toallen’s reagent) white solid of silver acetylide is separated.

Question 18.
Give four examples for isomerism in Arenes.
Answer:

Question 19.
Give three examples for fused polycyclic arene
Answer:

Question 20.
State Markownkoff s rule.
Answer: When an unsymmetrical molecule adds to an unsymmetrical alkene, the negative part of the adding molecule adds to the carbon atom involved in the double bond, containing the least number of hydrogen atoms. Thus according to Markonwkoff’s rule the addition takes place as follows :

1st PUC Chemistry Hydrocarbons Four/Five Mark Questions and Answers

Question 1.
Explain the action of chlorine on methane in the presence of sunlight.
Answer:
Methane reacts with chlorine in the presence of diffused sunlight to give methyl chloride, methylene chloride, chloroform and carbon tetrachloride.

Question 2.
Explain the mechanism of chlorination of methane, according to Free Radical.
Answer:
Mechanism of chlorination of methane involves three types.
1. Initiation : Chlorine absorbs energy and undergoes homolysis to give chlorine free radicals.
\(\mathrm{Cl}-\mathrm{Cl} \stackrel{\mathrm{hy}}{\longrightarrow} 2 \mathrm{Cl}^{\bullet}\)

Step 2 : Propagation : Chlorine free radical reacts with methane to give methyl free radical.
\(\mathrm{Cl}^{\bullet}+\mathrm{CH}_{4} \longrightarrow \mathrm{CH}_{3}^{\bullet}+\mathrm{HCl}\)
The methyl free radical reacts with chlorine to form methyl chloride and chlorine free radical.

Step 3 : Termination : Free radials combine to form stable products.

Question 3.
Explain the mechanism of addition of hydrogen bromide to propene.
Answer:
The mechanism of addition of hydrogen bromide to propene takes place in three steps. Step 1: Hydrogen bromide dissociates into H⁺ and Br-
H-Br → H⁺+Br^–

Step 2 : The electrophile H+ attacks propene to form carbocation.

of the two carbocations (I) and (II), carbocations (I) is more stable and is formed more readily.

Step 3 : The nucleophile Br attacks the carbocation (I) to give 2 – bromopropane.

Question 4.
How is the manufacture of LDPE & HDPE ?
Answer:
Low density polythene is manufactured by compressing ethane under very high pressure (1500-2000 atmospheres) at about 475 K in the presence of a small amount of oxygen as catalyst.

High density polythene is obtained by passing ethene under pressure of 6 atmospheres at about 340 K into heptane, an inert solvent containing Zielgler-Natta catalyst (triethyl) aluminium and titanium tetrachloride)

Question 5.
Give four uses of aceylene.
Answer:

1. Oxyacetylene flame is used for cutting and welding of metals.
2. Acetylene and its derivatives are widely used in synthetic organic chemistry for synthesis of cis and trans-alkenes, methyl ketones, etc.
3. Acetylene is used as illuminant in hawker’s lamp and in light houses.
4. Acetylene is used for ripening of fruits and vegetables.
5. Acetylene is used for manufacture of ethyl alcohol, acetaldehyde, acetic acid, vinyl plastics, synthetic rubbers such as Buna-N and synthetic fibres such as Orlan.

Question 6.
Give 4 uses of ethene.
Answer:

1. Lower members of the family are used as fuels and illuminants.
2. Alkenes and substituted alkenes upon polymerization form a number of useful polymers such as polythene, PVC, Teflon, orlon, etc.,
3. Ethene is employed for preparation of ethyl alcohol and ethylene glycol (anti-freeze).
4. Ethylene is used for artificial ripening of green fruits.
5. Ethylene also used in oxygen-ethylene flame for cutting and welding of metals.

Question 7.
Give four uses of methane.
Answer:

1. Methane in the form of natural gas (CNG) is used for running scooters, cars, buses, etc., LPG (Mixture of butane and isobutene) is used as a fuel in homes as well as in industry.
2. Methane is used to make carbon black which is used in the manufacture of printing inks, paints and automobile tyres.
3. Cataltic oxidation of alkanes gives alcohols, aldehydes an carboxylic acids.
4. Higher alkanes in form of gasoline, kerosene oil, diesel, lubricating oils and paraffin wax are widely used.
5. Methane is used for manufacture halogen containing compounds such as CH₂Cl₂ , CHCl₃ , CCl₄ etc. which are used as solvents both in laboratory and industry.

Question 8.
Give the structural elucidation of benzene.
Answer:
1. From elemental analysis and molecular mass determination, the molecular formula of benzene is found to be C₆H₆.

2. On the basis of molecular formula benzene is highly unsaturated compound.

3. Benzene does not show alkene properties. Benzene is quite stable. It does not decolourise cold aqueous solution of potassium paramanganate.

4. Benzene shows cyclic structure,
(a) Benzene contains three double bonds. Benzene adds three moles of hydrogen in presence of nickel catalyst to form cyclohexane.

\(\mathrm{C}_{6} \mathrm{H}_{6}+3 \mathrm{H}_{2} \frac{180^{\circ} \mathrm{C}}{\mathrm{Ni}} \mathrm{C}_{6} \mathrm{H}_{12}\)

(b) All the six hydrogen atoms in benzene are identical – benzene reacts with bromine in presence of FeBr3 catalyst to form monobromobenzene.

\(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \stackrel{\mathrm{FeBr}_{3}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr}\)

5. Kekule proposed benzene is a planar ring structure with alternate single and double bonds.

6. To overcome this drawback, Kekule suggested that benzene is a mixture of two forms, which are in rapid oscillation.

7. It is found that the bond length in benzene is same for all C-C bonds (0.139 nm). This lies between C-C single bond (0.154 nm) and C-C double bond length (0.134 nm).

8. Resonance structure: Resonance hybrid is more stable than structure 1 and 2. The stability of benzene due to resonance is so great that n bonds of the molecule will normally resist breaking. This explain the stability of benzene.

Question 9.
Explain the mechanism of halogenation or chlorination of benzene.
Answer:
Halogenation: Benzene reacts with chlorine in the presence of FeCl₃ or AlCl₃ to form chlorobenzene.

Machanism : This involves the following steps.

Step 1: Generation of eiectrohile Cl – Cl + FeCl₃ → Cl⁺ + \(\mathrm{FeCl}_{4}^{-}\)
Step 2 : The electrophile Cl⁺ attacks benzene ring to form a carbon cation which is resonance stabilised.

Step 3 : Loss of a proton to give chlorobenzene. The proton is removed by FeCU.

Question 10.
Explain the mechanism of nitration of benzene.
Answer:
Nitration benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric acid at 50°C to form nitrobenzene.

Mechanism : This involves the following steps.
Step 1: Generation of electrophile nitronium ion \(\mathrm{NO}_{2}^{+}\)
\(\mathrm{HNO}_{3}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NO}_{2}^{+}+\mathrm{H}_{3} \mathrm{O}^{+}+2 \mathrm{HSO}_{4}^{-}\)

Step 2 : The electrophile NO₂ attacks the benzene ring to form a carbocation which is resonance stabilized.

Step 3 : Loss of a proton to give nitrobenzene. The proton is removed by \(\mathrm{HSO}_{4}^{-}\)

Question 11.
Explain the mechanism of sulphonation of benzene?
Answer:
Sulphonation : Benzene reacts with concentrated sulphuric acid at 80°C to form benzene sulphoric acid.

Mechanism : This involves the following steps.

Step 1: Generation of electrophile SO₃.
\(2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{SO}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)

Step 2 : The electrophile SO₃ attacks the benzene ring to form a carbocation

Step 3: Loss of a proton to from a sulphonate ion. The proton is removed by \(\mathrm{HSO}_{4}^{-}\)

Step 4 : Addition of proton to give benzene sulphonic acid.

Question 12.
Explain the mechanism of Friedel craft alkylation of benzene.
Answer:
Friedel – Craft alkylation reaction Benzene reacts with methyl chloride in presence of anhydrous aluminium chloride to form toluene. This is called Friedel-Crafts alkylation reaction.

The mechanism of Friedel Craft’s alkylation reaction involves the following steps:

You can Download Chapter 16 Digestion and Absorption Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Question Bank Chapter 16 Digestion and Absorption

1st PUC Biology Digestion and Absorption NCERT Text Book Questions and Answers

Question 1.
Choose the correct answer among the following:
(a) Gastric juice contains
(i) Pepsin, lipase and rennin
(ii) Trypsin, lipase and rennin
(iii) Trypsin, pepsin and lipase
(iv) Trypsin, pepsin and renin
Answer:
(i) Pepsin, lipase and rennin

(b) Succus entericus is the name given to
(i) A junction between the ileum and large intestine
(ii) Intestinal juice
(iii) Swelling in the gut
(iv) Appendix
Answer:
(ii) Intestinal juice

Question 2.
Match column I with column II

Column I – Column II

(a) Bilirubin and biliverdin – (i) Parotid
(b) Hydrolysis of starch – (ii) Bile
(c) Digestion of fat – (iii) Lipases
(d) Salivary gland – (iv) Amylases
Answer:
(a) – ii
(b) – iv
(c) – iii
(d) – i

Question 3.
Answer briefly:
(1) Why are villi present in the intestine and not in the stomach?
Answer:
Because villi are supplied with a network of capillaries and a large lymph vessel called a lacteal. The mucosal epithelium has goblet cells which secrete mucus that help in lubrication. Mucosa also forms glands in the stomach (gastric glands) and crypts in between the bases of villi in the intestine (crypts of lieberkuhn).

(2) How does pepsinogen change into its active form?
Answer:
The proenzyme pepsinogen, on exposure to hydrochloric acid, gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteases and peptones (peptides).

(3) What are the basic layers of the wall of the alimentary canal?
Answer:
The wall of the alimentary from the esophagus to the rectum possesses four layers namely serosa, muscularity, submucosa, and mucosa. The serosa is the outermost layer and is made up of a thin mesothelium (epithelium of visceral organs) with some connective tissues. Muscularity is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. An oblique muscle layer may be present in some regions. The submucosal layer is formed of loose connective tissue containing nerves, blood, and lymph vessels. In the duodenum, glands are also present in the submucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa.

(4) How does bile help in the digestion of fats?
Answer:
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes. Bile helps in the emulsification of fats i.e. break down the fats into very small micelles. Bile also activates lipases. Fats are broken down by lipases with the help of bile into di-and monoglycerides.

Question 4.
State the role of pancreatic juice in the digestion of proteins.
Answer:
The pancreatic juice contains inactive enzymes trypsinogen, chymotrypsinogen, procarboxy¬peptidases, amylases, lipases, and nucleases. Trypsinogen is activated by an enzyme enterokinase secreted by intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. These enzymes are concerned with protein, carbohydrate, fats, and nucleic acid digestion.

Question 5.
Describe the process of digestion of protein in the stomach.
Answer:
The proenzyme pepsinogen present in the stomach gets converted to proteolytic enzyme pepsin in presence of Hydrochloric acid. Pepsin converts proteins into proteases and peptones. Rennin is another proteolytic enzyme present in the gastric juice of infants which helps in the digestion of milk proteins.

Question 6.
Give the dental formula of human beings.
Answer:

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Answer:
Bile is a watery greenish fluid mixture containing bile pigments, bile salts, cholesterol, and phospholipids. Bile helps in the emulsification of fats i.e. breaking down of the fats into smaller micelles, it also activates lipases. Thus, it is important for digestion.

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?
Answer:
Chymotrypsin is a protein digestive enzyme that breaks down proteins, peptones, and proteoses into dipeptides. The other two proteolytic enzymes are trypsin and carboxyl peptidase.

Question 9.
How are polysaccharides and disaccharides digested?
Answer:
About 30% of starch is hydrolysed by the enzyme salivary amylase into disaccharide maltose in oral cavity.

Carbohydrates in the chyme are hydrolysed by pancreatic amylase into disaccharides. Polysaccharides (starch)

Maltase present in the intestinal juice converts maltose into glucose. Lactase converts lactose into glucose and galactose. Sucrase converts sucrose into glucose and fructose

Question 10.
What would happen if HCl were not secreted in the stomach?
Answer:
Importance of HCl secreted by stomach:
HCl provides the acidic pH (pH 1.8) optimal for pepsins. It converts proenzyme pepsinogen into active enzyme pepsin, the proteolytic enzyme of the stomach.
HCl is also necessary to kill harmful bacteria which may be present in the food.

Question 11.
How does butter In your food get digested and absorbed in the body?
Answer:
Butter is rich in lipids. Digestion of fat starts in the stomach. The gastric lipase hydrolyses a small amount of lipids. Bile helps in the emulsification of fats and activates lipases which break down fats into diglycerides and then to monoglycerides. Monoglycerides and Diglycerides are further broken down to fatty acids and glycerol in presence of a lipase.

Fatty acids and glycerol being insoluble is converted to small droplets called micelles which are reformed to very small protein-coated fat globules called the chylomicrons which are absorbed by the intestine villi.

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Answer:
There is no protein digestion in the oral cavity as there are no proteases present.
Stomach: The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted to active enzyme pepsin. Which converts proteins into proteases and peptones.

Intestine: Pancreatic juice contains inactive enzymes like trypsinogen, chymotrypsinogen, and procarboxy peptidases. Trypsinogen is converted to active trypsin with the help of the enzyme enterokinase. Trypsin, in turn, activates other proteolytic enzymes in the pancreatic juice. Proteins, proteoses, and peptones in the intestine are now converted to dipeptides in presence of trypsin, chymotrypsin, and carboxypeptidase.

The aminopeptidase hydrolyses the peptide bond that attaches the terminal amino acid to the amino end of the peptide. Dipeptidase acts on dipeptides and converts them to amino acids.

Amino acids are the end products of protein digestion.

Question 13.
Explain the terms thecodont and diphyodont.
Answer:
The alimentary canal begins with an anterior opening-the mouth, and it opens out posteriorly through the anus. The mouth leads to the buccal cavity or oral cavity. The oral cavity has a number of teeth and a muscular tongue. Each tooth is embedded in a socket of the jaw bone. This type of attachment is called thecodont. The majority of mammals including humans forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Answer:
Incisors – 8, canine – 4, premolars – 8, molars – 12.

Question 15.
What are the functions of the liver?
Answer:
Functions of Liver:

1. The liver performs several roles in carbohydrate, lipid, and protein metabolism.
2. The liver is responsible for the mainstay of protein metabolism, synthesis as well as degradation.
3. The liver produces and excretes bile (a greenish liquid) required for emulsifying fats.
4. The breakdown of insulin and other hormones
5. The liver breaks down hemoglobin, creating metabolites that are added to bile as a pigment (bilirubin and biliverdin).
6. The liver converts ammonia to urea.

1st PUC Biology Digestion and Absorption Additional Questions and Answers

1st PUC Biology Digestion and Absorption One Mark Questions

Question 1.
What type of medium is required for the activity of trypsin?
Answer:
Alkaline medium.

Question 2.
Which cells secrete HCI?
Answer:
Parietal cells.

Question 3.
What is the other name for Pancreatic amylase?
Answer:
Amylopsin

Question 4.
What is chylomicron?
Answer:
The reconstructed triglycerides combine with phospholipids and cholesterol are released into the lymph in the form of protein-coated water-soluble fat globules or droplets. These are called chylomicron.

Question 5.
Which glands secrete HCI?
Answer:
Gastric glands.

Question 6.
Which digestive Juice is non-enzymatic?
Answer:
BWe.

Question 7.
Which gland secretes Bile?
Answer:
Liver.

Question 8.
Where is caecum located in the alimentary canals?
Answer:
The caecum is located at the junction of the small intestine and colon (large intestine).

Question 9.
What type of muscles present on the wall stomach?
Answer:
Smooth muscles.

Question 10.
Which part of small intestine is the longest?
Answer:
Ileum

Question 11.
Give an example of a vestigial organ.
Answer:
Vermiform appendix. (Oct. 83)

Question 12.
How is the tongue attached to the floor of the buccal cavity?
Answer:
The tongue is attached to the floor of the buccal cavity by the frenulum.

Question 13.
Mention the function of villi. (Oct. 84)
Answer:
The Villi are structures of the intestinal wall that increase the surface of absorption i.e. their chief function is absorption

Question 14.
What is ‘Bile’? (Oct. 85)
Answer:
Bile is the juice secreted by the liver.

Question 15.
What is the mechanical action taking place in the mouth called? (Oct. 90)
Answer:
The mechanical action taking place in the mouth is known as ‘Mastication’.

Question 16.
Name a proteolytic enzyme
Answer:
Pepsin (or trypsin, chymotrypsin, carboxypeptidase Aminopeptidase, tripeptidase, dipeptidase – anyone can be named)

Question 17.
Name the gland which is both exo and endocrine? (April 93)
Answer:
Pancreas.

Question 18.
What is the sphincter of Oddi?
Answer:
The sphincter of Oddi is the muscular structure, that guards the opening of the hepato-pancreatic duct into the duodenum.

Question 19.
Name the Enzyme that hydrolyses lipids.
Answer:
‘Lipases’ hydrolyze lipids. (Oct. 94)

Question 20.
Name the enzyme which does not act on the food materials in the small intestine.
Answer:
The “Enterokinase” enzyme.

Question 21.
What prevents the entry of food into the larynx during swallowing? (April 96)
Answer:
Epiglottis.

Question 22.
Which hormone causes Diabetes mellitus? (April 1996)
Answer:
The deficiency of ‘Insulin’ causes Diabetes mellitus.

Question 23.
Name the pancreatic hormone which acts as a hypoglycemic factor.(April 97, M.Q.P.)
Answer:
Insulin acts as a hypoglycemic factor.

Question 24.
Name the Carbohydrate digesting enzyme present in saliva. (Oct. 97)
Answer:
Salivary amylase or Ptyalin.

Question 25.
What is the basic mechanism of digestion in the intestine? (Oct. 88)
Answer:
‘Hydrolysis’.

Question 26.
The catalytic activity of which enzyme produces Fructose as one of the end products? (Oct. 88)
Answer:
Sucrase acts on the disaccharide sucrose resulting in a molecule of Fructose and a molecule of glucose.

Question 27.
What type of enzyme is pepsin? (Oct. 89)
Answer:
Proteolytic enzyme.

Question 28.
What Is emulsification? (Oct. 98)
Answer:
Emulsification is a process by which fats are converted to small water-soluble muscles by Bile salts making them easily accessible to the action of Fats digesting enzymes called lipases.

Question 29.
Mention the function of microvilli. (April 99)
Answer:
The chief function of the microvilli is absorption (i.e., it increases the surface of absorption).

Question 30.
What Is Chyme? (April 2000, July 2010)
Answer:
The liquified and acidified food present in the stomach is called as chyme.

Question 31.
What are the end products of protein digestion? (April 2001)
Answer:
Amino acids.

Question 32.
What is Peristalsis? (Oct. 2003)
Answer:
The powerful, rhythmic waves of muscular contraction and relaxation in the walls of hollow tubular organs like the digestive tract is called peristalsis.

Question 33.
Name an enzyme present in saliva meant for killing bacteria. (July 2006)
Answer:
Lysozyme.

Question 34.
What Is succus entericus? (April 2007)
Answer:
Succus entericus is the intestinal juice made of water, mucin and 7 digestive enzymes.

Question 35.
Give reason: (July 2008)
In the absence of Enterokinase, protein digestion is incomplete.
Answer:
Enterokinase activates trypsinogen to trypsin which breaks down proteins. Hence the absence of enterokinase results in indigestion of proteins.

Question 36.
Mention two characteristics of mammalian teeth.
Answer:
Heterodont and thecodont.

Question 37.
Why are proteases generally released In – inactive form? (All India 2003)
Answer:
proteases are inactive form, they would hydrolyse the cellular and extracellular proteins of the gut wall in the absence of food.

Question 38.
What are micelles? (All India 1998)
Answer:
The products of fat digestion are incorporated with the help of bile salts and phospholipids, into small spherical, water-soluble droplets called micelles.

Question 39.
Name the passage that leads bile from the liver Into gall bladder. (Delhi 1998)
Answer:
Cystic duct.

Question 40.
Name the different parts of large intestine in humans in their natural sequence. (Delhi 1998C)
Answer:
Caecum, colon and rectum.

Question 41.
What is the number of permanent teeth in an adult human being?
Answer:
32. (Thirty-two).

Question 42.
Name the two sets of teeth a human gets In his life.
Answer:
Milk teeth (deciduous teeth) and permanent teeth.

Question 43.
Name the substance that makes up the chewing surface of teeth.
Answer:
Ename

Question 44.
What are the four different types of teeth present in humans?
Answer:
lncisdrs, canine, premolars, and molars.

Question 45.
How many (I) molar and (II) premolar teeth are there in an adult human?
Answer:

• 12 molars
• 8 premolars.

Question 46.
Where are the taste buds located?
Answer:
Taste buds are located in the papillae on the upper surface of the tongue.

Question 47.
How is the tongue attached to the floor of the buccal cavity?
Answer:
The tongue is attached to the floor of the buccal cavity by the frenulum.

Question 48.
What is the function of epiglottis?
Answer:
Epiglottis prevents the entry of food into the trachea during swallowing.

Question 49.
Name the muscular structure that regulates the movement of food from the esophagus into the stomach.
Answer:
Gastro-Oesophageal sphincter.

Question 50.
Where is the stomach located?
Answer:
Stomach is located in the upper part of the abdominal cavity on the left side, just below the diaphragm.

Question 51.
Name the three regions of the stomach in proper sequence.
Answer:
Cardiac, fundic and pyloric regions.

Question 52.
Which part of the stomach does the oesophagus enter into?
Answer:
Cardiac region.

Question 53.
Which part of the stomach continues into the duodenum?
Answer:
Pyloric region.

Question 54.
Name the three regions of the human small intestine in the proper sequence.
Answer:
Duodenum, jejunum and ileum.

Question 55.
Name the muscular structure that guards the opening of the stomach Into the duodenum.
Answer:
Mr.-Pyloric sphincter.

Question 56.
Name the part of the alimentary canal where symbiotic microbes live.
Answer:
Caecum.

Question 57.
Name the three parts of the colon.
Answer:
Ascending, transverse and descending part.

Question 58.
Name any two structural features of the small intestine, which helps in absorption.
Answer:

• Presence of villi
• Projections called microvilli giving brush border appearance.

Question 59.
What is the name of the major lymph vessel present in the intestinal villi?
Answer:
Lacteal.

Question 60.
Where are the crypts of leiberkuhn located?
Answer:
They are located in between the bases of the villi in the intestine.

Question 61.
Name any two major glands associated with the human alimentary canal.
Answer:
Pancreas, liver, and the salivary glands.

Question 62.
Name the largest gland in our body.
Answer:
Liver.

Question 63.
Name the structural and functional unit of liver.
Answer:
Hepatic lobules.

Question 64.
Name the connective tissue sheath of liver lobules.
Answer:
Glisson’s capsule.

Question 65.
What are the constituents of the common bile duct?
Answer:
Hepatic duct and cystic duct.

Question 66.
How does bile reach the gall bladder?
Answer:
The bile secreted by the hepatic cells passes through the hepatic ducts and enters the gall bladder through the cystic duct.

Question 67.
What is sphincter of Oddi?
Answer:
Sphincter of Oddi is the muscular structure, that guards the opening of the hepato-pancreatic duct into the duodenum.

Question 68.
Name one gland in human body, that secretes both digestive enzymes as well as hormones.
Answer:
Pancreas.

Question 69.
What are the two major functions of buccal cavity?
Answer:
Buccal cavity helps in the mastication of food and facilitates in swallowing.

Question 70.
What is a bolus?
Answer:
When thoroughly masticated food mixes with saliva, the food particles stick together with the help of mucus into what is known as bolus.

Question 71.
Define deglutition.
Answer:
Deglutition is the process of swallowing in which bolus is conveyed into the pharynx and then into the oesophagus.

Question 72.
In humans, starch digestion begins in the buccal cavity, but stops in the stomach. Why?
Answer:
Hydrochloric acid in the gastric juice (acidic pH) inactivates salivary amylase. Hence starch digestion stops in the stomach.

Question 73.
What is the role of the intrinsic factor?
Answer:
lntrinsic factor helps in the absorption of vitamin B₁₂ from the intestine.

Question 74.
Name two enzymes which take part in the digestion of proteins in our body.
Answer:
Pepsin, trypsin, chymotrypsin and carboxypeptidase.

Question 75.
What type of medium (pH) is required for the activity of trypsin?
Answer:
Trypsin requires an alkaline pH.

Question 76.
What is the role of intrinsic factor?
Answer:
Intrinsic factor helps in absorption of vitamin B12from the intestine.

Question 77.
Name two enzymes which take part in the digestion of proteins in our body.
Answer:
Pepsin, Trypsin, Chymotrypsin and carboxypeptidase.

Question 78.
What type of medium (pH) is required for the activity of trypsin?
Answer:
Trypsin requires an alkaline pH.

Question 79.
What provides the alkaline medium for the action of trypsin in small Intestine?
Answer:
Alkaline medium is provided by bile from the liver and bicarbonates from the pancreas and Brunner’s glands.

Question 80.
What are peptones?
Answer:
Peptones are partially hydrolysed proteins.

Question 81.
Which is the food constituent which bile helps to digest and absorb?
Answer:
Fats.

Question 82.
Name the bile pigments?
Answer:
Bilirubin and biliverdin.

Question 83.
Name the end products of digestion of fats.
Answer:
Glycerol, fatty acids, and monoglycerides.

Question 84.
Define absorption.
Answer:
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into the blood or lymph.

Question 85.
Name the form in which fats enter lymph.
Answer:
Chylomicrons.

Question 86.
What are chylomicrons?
Answer:
Chylomicrons are the protein-coated, water-soluble globules of the newly synthesized fats that is combined with phospholipids and liberated into the lymph for circulation.

Question 87.
What is defaecation?
Answer:
The egestion of faeces through the anal opening is called defaecation.

Question 88.
Name the most common disorder of the alimentary system.
Answer:
Inflammation of the intestinal tract.

Question 89.
Which of the body is affected by jaundice?
Answer:
Liver.

Question 90.
What is Constipation?
Answer:
Constipation is the condition where the faeces are retained within the rectum, as the bowel movement occurs irregularly.

Question 91.
What is facilitated transport?
Answer:
Some of the substances like fructose and few amino acids are absorbed by the blood with the help of carrier ions like Na+. This mechanism is called facilitated transport.

1st PUC Biology Digestion and Absorption Two Marks Questions

Question 1.
If the pancreatic duct of a person is blocked, how would it affect the digestion of fat in the duodenum?
Answer:
If the pancreatic duct is blocked, the pancreatic juice cannot reach the duodenum. As a result, the enzymes like trypsin, chymotrypsin, carboxypeptidase, aminopeptidase (help in the digestion of protein); amylase (help in carbohydrate digestion), lipase (help in fat digestion) could not reach the duodenum. Digestion of carbohydrates, proteins, and fats will remain incomplete.

Question 2.
Mention any four proteolytic enzymes Involved in human digestion. (April 90)
Answer:

• Pepsin
• Trypsin
• Chymotrypsin
• Carboxypeptidase.

Question 3.
Name the watery fluid secreted from Brunner’s gland in the duodenum. Mention its any two characteristics. What role does it play inside the duodenum?
Answer:
The watery fluid secreted by Brunner’s gland is called mucoid fluid or mucus.
Characteristics:
(i) It is viscous.
(ii) It is enzyme-free
(iii) It is alkaline in nature.
Functions:
It enables the duodenum to withstand the acidic chyme entering from the stomach.

Question 4.
Write the role of Trypsin and Chymotrypsin in protein digestion. (April 2002)
Answer:
Trypsin is secreted as trypsinogen. It is activated by the enterokinase of the intestinal juice.

Trypsin breaks down proteins into proteases, peptones & polypeptides.

polypeptides.Chymotrypsin is secreted as chymotrypsinogen. It is activated by trypsin. It too converts proteins into protein fragments, Chymotrypsinogen

Question 5.
Mention the three types of salivary glands in man and name the carbohydrate present in the saliva. (Oct. 2002)
Answer:
In man there are 3 pairs of salivary glands, they are are the parotid, Sublingual and Submaxillary glands. Saliva contains the starch splitting enzyme salivary amylase or Ptyalin.

Question 6.
Name the non-digestive enzyme found in intestinal juice. Mention its significance. (Oct. 2004)
Answer:
The non-digestive enzyme in intestinal juice is ‘E-‘nterokinase’ and helps to activate trypsinogen into trypsin which digests proteins.

Question 7.
What are microvilli? State their functions.
Answer:
Microvilli are bristle-like extensions of the free surface of epithelial cells that line the surface of villi. They increase the surface area of epithelium for the absorption of nutrients.

Question 8.
Write any two roles of HCI In human digestion. (July 2010)
Answer:
The major roles played by HCI in human digestion are;

• It activates pepsinogen to pepsin which is required for protein digestion.
• It prevents decay of food in the stomach.

Question 9.
How is our gut lining protected from its own secretion?
Answer:

• Proteases are secreted in an inactive form and pose no threat to the gut lining.
• The mucus and bicarbonates present in the gastric juice play an important role in the protection of the mucosal epithelium from excoriation by the highly concentrated Hydrochloric acid.

Question 10.
What would happen if hydrochloric acid is not secreted in our stomach?
Answer:
If hydrochloric acid is not secreted in the stomach, the following will happen:

• Pepsinogen will not be converted into pepsin.
• An acidic medium needed for the action of proteases will not be created.
• Salivary amylase may continue to function.

Question 11.
Amylase is secreted by two different glands. Name them what is its action on food? (Foreign 2002)
Answer:
Amylase is secreted by the salivary glands into the buccal cavity and also by the pancreas.
Amylase acts on starch and breaks it into two molecules of glucose.

Question 12.
What is the site of fat digestion In humans? Name the enzymes responsible for it and mention its end products.
Answer:
Fat is mainly digested in the small intestine and very little in the stomach.
Lipase is the enzyme responsible for fat digestion. The final products of fat digestion in humans are glycerol, fatty acids and some monoglycerides.

Question 13.
Enumerate the contents of saliva.
Answer:
The saliva secreted into the oral cavity contains electrolytes like Na⁺, K⁺, Cl^–, \({ HCO }_{ 3 }^{ – }\) and enzymes like salivary amylase and lysozyme. Salivary amylase helps in digesting starch and lysozyme acts as an antibacterial agent that prevents infections.

Question 14.
What is rennin? What is its function?
Answer:

• Rennin is the proteolytic enzyme found in the gastric juice of infants.
• It hydrolyses milk proteins and helps in its digestion.

Question 15.
How would it affect the digestion of proteins and carbohydrates, if there is a blockage in the pancreatic duct?
Answer:
If the pancreatic duct is blocked:

• Digestion of proteins will be impaired, as pancreatic juice contains the proteolytic enzymes trypsin, chymotrypsin and carboxypeptidase.
• Digestion of carbohydrates will stop as there will be no enzyme amylase.
• Since the bicarbonates provide an alkaline pH for the action of enzymes in the duodenum, the whole process of digestion in the intestine will be affected.

Question 16.
Give two protein-digesting enzymes of pancreatic Juice with their action.
Answer:
Pancreatic juice contains enzymes like trypsin, chymotrypsin and carboxyptidase which help in digestion of proteins.

• Trypsin acts on the proteins, proteoses and peptones and convert them into small peptides/ dipeptides
• Carboxypeptidase release the last amino acid from the peptide chain and shorten the peptide chain.

Question 17.
What are the basic layers of the wall of the alimentary canal?
Answer:
The wall of the alimentary canal consists of four main concentric layers. Beginning from outside, these layers are

• visceral peritoneum
• muscular layer
• sub-mucosa
• mucosa.

Question 18.
Mention the Important functions of the large intestine.
Answer:

• It helps in the absorption of water, minerals, and certain drugs.
• Secretion of mucus helps in adhering the waste particles together and lubricating it for an easy passage.

Question 19.
Where is the ileocaecal valve present? What are its functions?
Answer:

• The ileo-caecal valve is present at the junction of ileum of small intestine and the caecum of the large intestine.
• It prevents the backflow of the matter from the caecum into the ileum.

Question 20.
What Is diarrhea? What is its consequence In nutrition?
Answer:
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhea. It reduces the absorption of food.

Question 21.
Briefly explain the structure and regions of the stomach.
Answer:
Stomach is a ‘J’ shaped bag-like structure. A muscular sphincter called ‘gastro-oesophageal sphincter regulates the opening of oesophagus into the stomach. The stomach is located in the upper-left portion of the abdominal cavity and has three major parts a cadiac portion into which the oesophagus opens, a fun dic region and a pyloric region which opens into the small intestine.

Question 22.
How Is saliva produced?
Answer:
Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub-mandibular and the sub linguals. These glands are situated just outside the buccal cavity and secrete salivary juice into the buccal cavity.

Question 23.
Where is the pancreas situated? Name the secretions of the pancreas.
Answer:
The pancreas is a compound (both exocrine and endocrine) elongated organ situated between the limbs of the U shaped duodenum. The exocrine portion of the pancreas secretes an alkaline pancreatic juice containing enzymes liketrypsinogen, chymotrypsinogen, procarboxy peptidases, amylases, lipases and nucleases. The endocrine portion secretes hormones, insulin and glucagon.

Question 24.
What constitutes the juice? What does it .contain?
Answer:
The secretions of the brush border cells of the mucosa along with the secretions of the gobletcells constitute the intestinal juice or succus entericus. This juice contains a variety of enzymes like disaccharidases (e.g. maltase) dipeptidases, lipases and nucleosidase.

Question 25.
How are fatty acids and glycerol absorbed into the blood?
Answer:
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. They are re-formed into very small protein-coated fat globules called the chylomicrons which are transported into the lymph vessels in the villi. These lymph vessels ultimately release the absorbed substances into the bloodstream.

Question 26.
Draw a labelled diagram of anatomical regions of the human stomach.
Answer:

1st PUC Biology Digestion and Absorption Three Marks Questions

Question 1.
Draw a neat labelled diagram representing the duct system of liver, gall bladder and pancreas.
Answer

Question 2.
Name any three enzymes secreted by the pancreas Specify the substrate and product of each.
Answer:

• Amylase: It acts on starch and converts it into maltose.
• Trypsin: It acts on proteins, proteases and peptones and converts them into shorter peptides or dipeptides.
• Lipase: It acts on triglycerides and converts them into triglycerides and monoglycerides along with the release of fatty acids.

Question 3.
Draw a neat labelled diagram representing transverse section of gut.
Answer:

Question 4.
What is protein-energy malnutrition? How it is responsible to cause marasmus and kwashiorkor in infants and children?
Answer:
Protein-energy malnutrition (PEM) may affect large sections of the population during drought and famine.
This happened in Ethiopia during the severe drought in the mid-eighties. PEM affects infants and children to produce Marasmus and Kwashiorkor diseases.

• Marasmus is produced by a simultaneous deficiency of proteins and calories. It is found in infants less than a year in age if mother’s milk is replaced too early by other foods which are poor in both proteins and caloric value.
• This often happens if the mother has second pregnancy or childbirth when the older infant is still too young.
• In Marasmus, protein deficiency impairs growth and replacement of tissue proteins; extreme emaciation of the body and thinning of limbs results, the skin becomes dry, thin, and wrinkled.
• Growth rate and bodyweight decline considerably. Even growth and development of the brain and mental abilities are impaired.
• Kwashiorkor is proceeded by protein deficiency unaccompanied by calorie deficiency.
• It resubs from the replacement of mother’s milk by a high calorie-low protein diet in a child more than one year in age. Like marasmus, kwashiorkor shows wasting of muscles, thinning of limbs, failure of growth, and brain development.
• But unlike, marasmus, some fat is still left under the skin; moreover, extensive oedema and swelling of body parts are seen.

Question 5.
Draw a neat labelled diagram of a section of small intestinate mucosa showing villi.
Answer:

Question 6.
Which layer of the stomach contains gastric glands? Name the three types of cells present in gastric glands with their functions.
Answer:
The mucosa of the stomach has gastric glands.
Three major types of cells in gastric glands are:

• Mucus neck cells which secrete mucus
• Peptic or chief cells secrete the proenzyme pepsinogen.
• parietal or oxyntic cells which secrete HCL and intrinsic factors essential for the absorption of vitamin B_12.

Question 7.
How are the nucleic acid fraction of our food digested?
Answer:
Nucleases present in the pancreatic juice convert the nucleic acids to nucleotides.

• Nucleotidases present in the pancreatic juice convert the nucleotides into nucleosides and phosphates.
• Nucleosides present in the intestinal juice convert the nucleosides into nitrogen bases and sugar.

Question 8.
Write a note on hormonal control and co-ordination of digestive parts.
Answer:
The activities of the gastrointestinal tract are under neural and hormonal control for proper co-ordination of different parts. The sight, smell, or the presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also, similarly, stimulated by neural signals. The muscular activities of different parts of the alimentary canal can also be moderated by neural mechanisms, both local and through CNS. Hormonal control of the secretion of digestive juices is carried out by the local hormones produced by the gastric and intestinal mucosa.

Question 9.
Write a note on absorption in different parts of the digestive system
Answer:
Mouth: Certain drugs coming in contact with the mucosa of the mouth and lower side of the tongue are absorbed into the blood capillaries lining them.

Stomach: Absorption of water, simple sugars, and alcohol, etc. takes place in the stomach. Small intestine: Principle organ for the absorption of nutrients. The digestion is completed here and the final products of digestion such as glucose, fructose, fatty acids, glycerol, and amino acids are absorbed through the mucosa into all bloodstream and lymph.

Large Intestine: Absorption of water, some minerals, and drugs take place in the large intestine.

1st PUC Biology Digestion and Absorption Five Marks Questions

Question 1.
Mention the components of human saliva. Give the function. (M.Q.P.)
Answer:
The salivary glands produce a secretion called salivary juice or saliva. It is composed
Nucleosidases of water; mucin; ions like sodium, chlorides, HCO₃ etc; a bactericidal enzyme and a digestive enzyme called ptyalin or salivary amylase.
Functions of Saliva

• Moistens dry food and helps in swallowing
• Keeps mouth and teeth clean.
• Helps in tasting food
• The Salivary amylase helps in digesting the polysaccharides, starch & glycogen to maltose, a disaccharide.
• The bactericidal enzyme kills microbes and checks their load in food.
• Many substances like Iodides, lead etc are excreted through saliva.

Question 2.
Describe the digestion of carbohydrates in the small intestine. (Apr. 83, Oct. 04)
Answer:
The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase, and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

Question 3.
Explain protein digestion In small Intestine? (M.Q.P., Apr. 83, Oct. 99, Mar. 2010, July 2011)
Answer:
The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 4.
Describe the digestive process in the stomach and small intestine of man. (April 84)
Answer:
Digestion in the stomach: The chief organic molecules in food that are digested in the stomach are the proteins (native) and short chain simple lipids. The cells of the mucosal wall produce the gastric juice and it consists of the following enzymes apart from HCI and mucin.

• Pepsin- proteolytic enzyme(endopeptidase)
• Renin – milk curdling enzyme (Calf and infants of humans)
• Gastric lipase – acts on simple lipids. Pepsin acts on native protein molecules of food in the lumen of the stomach converting them into proteoses, peptones, and polypeptides.

In Infants & calf, where the chief source of food is milk, renin is an additional enzyme which helps in the curdling of milk protein casein, and later this curdled casein is digested by the action of pepsin and converted to proteoses, peptones and polypeptides.

proteoses, peptones, and polypeptides. Digestion in Small Intestine: Two types of juices are secreted into the intestinal lumen namely the pancreatic juice and intestinal juice. Both their juices possess enzymes digesting all types of complex organic molecules namely carbohydrates, proteins, lipids & nucleic acids.
The digestion by this enzyme in the small intestine proceeds as follows:

Carbohydrates: Polysaccharides starch (plant source) and glycogen (animal source) are acted upon by pancreatic amylase (amylopsin) in the small intestinal lumen and converted to a pool of maltose, a disaccharide.

The disaccharides, in the intestinal lumen namely maltose (obtained from the digestion of polysaccharides or direct source); sucrose, and Lactose (from diet) are acted upon disaccharide digesting (disaccharidases) enzyme into a pool of monosaccharides, glucose, fructose, and galactose.


Proteins: The proteins reaching the small intestinal lumen are in the form of native proteins, proteoses, peptones, and polypeptides. They are acted upon by the following enzymes.

Thus resulting in a final pool of absorbable amino acids.

Lipids: The Lipids are first emulsified by bile salts in the intestinal lumen. The emulsified lipids are then worked upon by the pancreatic lipase to a pool of glycerol, fatty acids and monoglycerides.

Fattyacids, monoglycerides

Nucleic acids: Nucleases from the pancreas” acts upon the RNA & DNA nucleic acids breaking them into nucleotides and these nucleotides are later acted upon by nucleotidases & nucleosidases of the intestinal juice.


deoxyribose sugar; ribose sugar; nucleic acid bases.

Question 5.
Are mention two enzymes present in pancreatic juice? Explain their role In intestinal digestion. (April 86)
Answer:
The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralized by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotrypsin and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and aminopeptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 6.
Explain the digestion of food in the small intestine by the enzymes of intestinal glands. (Oct. 87)
Answer:
The secretion produced by the intestinal glands is known as the intestinal juice or succus entries. It is a clear yellow fluid with a pH of around 7.6 and contains water, mucous and some digestive enzymes produced by intestinal cells. The enzymes found in the intestinal juice and their action is as follows on food in the small intestine.

(1) Carbohydrases: Maltase, Lactase, Sucrase these are disaccharides digesting disaccharides Maltose, Lactose, and sucrose in the intestinal lumen as follows at pH 7.1 – 8.2.

(2) Proteases: Aminopeptidase, tripeptidase, and dipeptidase-(Exopeptdases) are proteolytic enzymes digesting the intermediate polypeptides produced by the action of endopeptidases of the stomach & pancreas. Their action is as follows in the small intestinal at pH 7.1 – 8.2


(Enterokinase, an activator enzyme, activates gastric glands to produce gastric juice)

(3) Lipase: A fat-digesting enzyme. It is not as strong as pancreatic lipase but it completes the digestion of fat in the small intestinal lumen.

(4) Nucleases: Act on nucleoside molecules of nucleoproteins, splitting them into corresponding nitrogenous bases and sugar molecules.

Question 7.
Explain the role of the pancreas in the process of digestion. (Oct. 87, 88)
Answer:
The pancreas is a dual gland composed of two portions, an exocrine portion having cells producing enzymes required for digestion of food and an Endocrine portion having groups of cells producing hormones required for metabolism and homeostasis.

As a digestive gland, it plays a very important role in the process of digestion by possessing the most potent or powerful enzymes required for digestion of all types of food molecules namely carbohydrates, proteins, fats, and nucleic acids.

The pancreatic juice secreted by the pancreatic exocrine glands reaches the intestine through the pancreatic duct (duct of wiring) which usually joins the bile duct before opening into the first part of the duodenum. It is a clear, colourless liquid, containing some salts and sodium bicarbonate apart from enzymes.

As soon as food enters the duodenum, a series of neurogenic and hormonal mechanism’s come into play and maintain a steady secretion of these fluids into the duodenum. These secretions mix with the acid chyme released from the stomach’ in the duodenum. The salts and bicarbonates neutralize and alkalize the acid chyme whereas the enzymes work upon the complex food molecules converting them into smaller molecules which are later on acted upon by intestinal juice enzymes converting them into absorbable forms.

The pancreatic digestive action can be summarized as follows :

(1) Carbohydrases (carbohydrate splitting): Pancreatic amylase acts on polysaccharides starch (plant origin) and glycogen (animal origin) converting these into disaccharide molecules, namely maltose.

(2) Proteases (Protein-splitting):

• Trypsin, Chymotrypsin (Endopeptidases) act on native proteins splitting them into proteases, peptones, and polypeptide molecules.
• Carboxypeptidase (exopeptidase) acts on proteins, peptones, and polypeptides splitting them into tripeptides, dipeptides, and amino acids.

(3) Pancreatic lipase: act on fats emulsified by brie salts splitting them into glycerol, fatty acids, and monoglycerides.

(4) Nucleases: Ribonuclease and deoxyribonuclease – splitting nucleic acids (RNA & DNA) into simple nucleotides of respective types. These are later acted upon by nucleotides, splitting them into nucleosides of RNA & DNA and PO

Question 8.
Explain how carbohydrates and protein are digested by intestinal juice. (Oct. 90)
Answer:
The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryrin, and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carboxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 9.
Explain the Digestion of protein In the human digestive system. (Oct. 92)
Answer:
Digestion in the stomach: The chief organic molecules in food that are digested in the stomach are the proteins (native) and short-chain simple lipids. The cells of the mucosal wall produce the gastric juice and it consists of the following enzymes apart from HCI and mucin.

• Pepsin- proteolytic enzyme(endopeptidase)
• Renin – milk curdling enzyme (Calf and infants of humans)
• Gastric lipase – acts on simple lipids. Pepsin acts on native protein molecules of food in the lumen of the stomach converting it into proteoses, peptones and polypeptides.

In Infants & calf, where the chief source of food is milk, renin is an additional enzyme which helps in the curdling of milk protein casein and later this curdled casein is digested by the action of pepsin and converted to proteoses, peptones and polypeptides.

proteoses, peptones and polypeptides. Digestion in Small Intestine: Two types of juices are secreted into the intestinal lumen namely the pancreatic juice and intestinal juice. Both their juices possess enzymes digesting all types of complex organic molecules namely carbohydrates, proteins, lipids & nucleic acids.
The digestion by this enzyme in the small intestine proceeds as follows:

Carbohydrates: Polysaccharides starch (plant source) and glycogen (animal source) are acted upon by pancreatic amylase (amylopsin) in the small intestinal lumen and converted to a pool of maltose, a disaccharide.

The disaccharides, in the intestinal lumen namely maltose (obtained from the digestion of polysaccharides or direct source); sucrose and Lactose (from diet) are acted upon disaccharide digesting (disaccharidases) enzyme into a pool of monosaccharides, glucose, fructose and galactose.


Proteins: The proteins reaching the small intestinal lumen are in the form of native proteins, proteoses, peptones and polypeptides. They are acted upon by the following enzymes.

Thus resulting in a final pool of absorbable amino acids.

Lipids: The Lipids are first emulsified by bile salts in the intestinal lumen. The emulsified lipids are then worked upon by the pancreatic lipase to a pool of glycerol, fatty acids, and monoglycerides.

Fatty acids, monoglycerides

Nucleic acids: Nucleases from the pancreas” acts upon the RNA & DNA nucleic acids breaking them into nucleotides and these nucleotides are later acted upon by nucleotidases & nucleosidases of the intestinal juice.


deoxyribose sugar; ribose sugar; nucleic acid bases.

(ii) The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and aminopeptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 10.
What is digestion? Explain the physiology of digestion of carbohydrates and protein In the small intestine. (Oct. 94, March 2011)
Answer:
Digestion is a process by which complex organic food molecules are broken down or hydrolysed into simple (or small) absorbable forms by the action of digestive enzymes.

The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by the Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 11.
Give an account of the digestion of proteins in man. (Oct. 2001)
Answer:
Protein digestion in man:
(a) Protein digestion in the mouth: Proteins do not undergo any change as there is no protease in the mouth.
(b) Protein digestion in the stomach: Gastric juice has two proteases pepsin and rennin.

(i) Pepsin is first secreted as pepsinogen: It is converted to pepsin by HCI. One HCI has converted some pepsinogen, further conversion is autocatalytic. Pepsin is an endopeptidase. It acts on internally situated peptide bonds of proteins. In the end, the proteins are converted to proteoses, peptones and polypeptides.

(ii) Rennin (also called chymosin): It is a milk curdling enzyme that is mostly present in infants. It is secreted as prorennin. Renin acts on the soluble milk protein, casein, converting it into paracasein. In the presence of Ca²⁺ of milk, it is further converted into calcium para caseinate, which coagulates preventing the rapid passage of milk into the duodenum. It is then hydrolyzed by pepsin to proteoses, peptones & polypeptides

As digestion proceeds in the stomach, the food becomes more or less liquified and is referred to as chyme.

(c) Protein digestion in the Intestine: Here it comes under the action of
(i) Pancreatic proteases &
(ii) Intestinal proteases.

(i) Pancreatic proteases: These are Trypsin, chymotrypsin, and carboxypeptidase. The first two are endopeptidases and the third is an exopeptidase. Trypsin is secreted as trypsinogen & it is activated by enterokinase (enteropeptidase) of the intestinal juice.

chymotrypsinogen which is activated by trypsin. It, too, converts proteins into protein fragments.

Carboxypeptidase an endopeptidase, hydrolyzes terminally situated bonds. The products are tripeptides, Depeptides, and amino acids.

(ii) Intestinal proteases: They may be termed peptidases because they act on the protein fragments. Proteoses, peptones, and polypeptides aminopeptidase tripeptides, dipeptides and

The proteins have been completed digested and the products are free amino acids. They are now ready to be absorbed.

Question 12.
Draw a neat labelled diagram of the human alimentary canal. (M.Q.P.)
OR
Draw a neat labelled diagram of digestive
Answer:

Question 13.
Write the functions of intestinal juice.
Answer:

• Intestinal mucus protects the intestinal epithelia.
• Disaccharidases like maltase, lactase, and sucrase bring complete hydrolysis of disaccharides into monosaccharides (see carbohydrate digestion).
• Aminopeptidases hydrolyse peptides of different lengths into amino acids.
• Dipeptidase hydrolyse dipeptides into amino acids, (see protein digestion).
• Nucleotidase and Nucleosidase hydrolyse Nucleotides and convert them into purines, pyrimidine, and inorganic phosphate.

Question 14.
Describe the process of Lipid digestion in small intestine of man. (Oct. 2004)
Answer:
Fats or lipids mainly contain triglycerides i.e. consists of 3 fatty acid molecules attached to a molecule of glycerol. The major fat-digesting enzyme is pancreatic lipase.

The fats are first emulsified i.e. the bile salts combine with fats and break them down into small droplets which form a soapy mixture. The small droplets are acted by lipase which breaks them into fatty acids, glycerol, mono and diglycerides.

The bile salts then combine with the products and form water-soluble complexes which are absorbed.

Question 15.
Briefly explain the disorders of the digestive system.
Answer:
The disorder of the digestive system are:-

• The inflammation of the intestinal tract is the most common ailment due to bacterial or viral infections. The infections are also caused by the parasites of the intestine like tapeworm, roundworm, threadworm, hookworm, pinworm etc.
• Jaundice:- The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.
• Vomiting: It is the ejection of stomach contents through the mouth. This reflex action is controlled by the vomiting centre in the medulla. A feeling of nausea precedes vomiting.
• Diarrhoea: The abnormal frequency of bowel movement and increased liquidity of the faecet discharge is known as diarrhoea. It reduces the absorption of food.
• Constipation: In constipation, the faeces are retained with the rectum as the bowel movements occur irregularly.
• Indigestion: In this condition, the food is not properly digested leading to a feeling of fullness. The causes of indigestion are inadequate enzyme secretion, anxiety food poisoning, overeating, and spicy food.

Question 16.
Fill in the blanks.

(a) The type of dentition in humans is called ……………..
(b) ……………..has …………….. cells which secrete mucus that help In lubrication.
(c) Each lobule of liver is covered by a thin connective tissue sheath called ……………..
(d) Hepate pancreatic duct is guarded by ……………..
(e) Alcohol is absorbed in ……………..
Answer:
(a) Diphyodont
(b) Mucosal epithelium, goblet cells
(c) Glisson’s capsule
(d) Sphincter of Oddi
(e) Stomach.

Question 17.
Match the following:
(a) Trypslnogen – (i) Goblet cells
(b) Saliva – (ii) Oxyntic cells
(c) HCL- (iii) Pancreas
(d) Pepsinogen – (iv) Parotids
(e) Mucus – (v) Chief cells
Answer:
(a) → (iii)
(b) → (iv)
(c) → (ii)
(d)→ (v)
(e) → (i)

You can Download Chapter 2 Structure of Atom Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

1st PUC Chemistry Structure of Atom One Mark Questions and Answers

Question 1.
Mention the constituents of atom.
Answer:
Electron, Proton & Neutron.

Question 2.
Who discovered electron.
Answer:
J.J. Thomson

Question 3.
What is the mass of electron?
Answer:
9.108 × 10³¹ kg (0.0005487amu).

Question 4.
Mention the charge of electron in colomb.
Answer:
1.61.6*10¹⁹C.

Question 5.
What is the charge of proton ?
Answer:
Positive charge.

Question 6.
What is the mass of proton ?
Answer:
1.672 x 10²⁷ kg(1.00728amu)

Question 7.
Who discovered proton ?
Answer:
E. Goldstein.

Question 8.
Who invented charge on electron ?
Answer:
Robert Millikan.

Question 9.
Name the experiment by which charge on electron determined.
Answer:
Robert Millikan oil drop experiment.

Question 10.
Who discovered neutron ?
Answer:
James Chadwick.

Question 11.
What is atomic number ?
Answer:
Atomic number *Z’ is the number of protons present in the nucleus of an atom of an element or the number of electrons present in an atom of an element.

Question 12.
What is mass number ?
Answer:
Atomic mass ‘A’. The total number of protons and neutrons present in the nucleus of an atom of an element.

Question 13.
Write the relationship between mass number and atomic number.
Answer:
Number of neutrons = Mass number (A) – Atomic number (Z) = A – Z

Question 14.
How do you represent an atom symbolically with atomic number and mass number?
Answer:
_ZX^A where A = Mass Number, X = atom, Z = Atomic number.

Question 15.
What is the number of proton and neutron in ₉₂X²³⁵ ?
Answer:

Question 16.
Give the number of Protons, Electrons and Neutrons present in the atom – having atomic number 27 and mass number 56.
Answer:
Number of Protons 27, Number of Electrons 27, Number of Neturons 29.

Question 17.
Mention the proton, neutron and electons ₁₇C³⁵.
Answer:
Neutron = 35 – 17 = 18, Electron = Proton = 17 .

Question 18.
Name the species which has no electron.
Answer:
Proton

Question 19.
Name the atom which has no neutron.
Answer:
Hydrogen

Question 20.
What is the ratio between mass of proton and electron ?
Answer:
Mass of proton is 1837 times the mass of electron.

Question 21.
Name the particles which constitute cathode rays.
Answer:
Electrons

Question 22.
Who demonstrated the particle property of an electron ?
Answer:
J. J. Thomson

Question 23.
Who showed the wave property of electrons ?
Answer:
G.P. Thomson

Question 24.
What is the charge on neutrons ?
Answer:
Nil / Zero

Question 25.
Mention the mass of neutron.
Answer:
1.675 × 10²⁷ kg(1.00866 amu)

Question 26.
Name the scientist who proposed the nuclear theory (Solar model of atom) theory and discovered the existence of nucleus of an atom ?
Answer:
Ernest Rutherford.

Question 27.
What is photon ?
Answer:
Electromagnetic radiations are emitted or absorbed or propagated in the form of small pockets of energy called photon.

Question 28.
What is Isotopes ?
Answer:
Atoms of the same element having the same atomic number but different mass numbers are called isotopes.

Question 29.
What is Isobars ?
Answer:
Atoms of different elements having the same mass number but different atomic numbers are called isobars.

Question 30.
What is Isotones ?
Answer:
Atoms of different elements which contain the same number of neutrons, but different mass numbers are known as isotones.

Question 31.
Define the term ‘radiation’
Answer:
Radiation is the emission and transmission of energy through space in the form of waves.

Question 32.
How are velocity, frequency and wavelength of light radiation related ?
Answer:

Question 33.
State Pauli’s exclusion principles.
Answer:
No two electrons in an atom can have the same set of four quantum numbers.

Question 34.
An atom having mass number 40 has 20 neutrons in its nucleus. What is the atomic number of the element ?
Answer:
Mass Number = Number of protons + Number of neutrons
40 = Number of protons + 20
∴Atomic number = Number of protons = 20.

Question 35.
What is aufbau principle ?
Answer:
Aufbau principle is also known as the building up principle. According to this principle, electrons should eneter the subshells in the order of increasing energies.

Question 36.
What is emission spectrum ?
Answer:
The spectrum of radiation emitted by a substance that has absorbed energy is called in emission spectrum.

Question 37.
How is the magnetic moment of paramagnetic species is related to the number of unpaired electrons present in it ?
Answer:
Magnetic moment \(\mu=\sqrt{n(n+2) B M}\) where n is the number of unpaired electrons.

Question 38.
Differentiate between the terms ‘ground state’ and ‘excited state’.
Answer:
Ground state means the least energy state or the most stable state. Excited state means the higher energy state, in which the electrons are in the higher energy level (unstable state).

Question 39.
What is the expression for the energy of a photon ?
Answer:
E = hu where, h = planks constant = 6.626 × 10^-34 Js , γ = frequency or radiation.

Question 40.
Write the unit for frequency of radiation.
Answer:
Cycles per sec (sec^-1) or Hertz (Hz)

Question 41.
Name the experiment which shows that light has particle property.
Answer:
Photoelectric effect.

Question 42.
Name the experiment which shows that light has wave property.
Answer:
Refraction.

Question 43.
How is wave number and wavelength of a wave related?
Answer:

Question 44.
What is the velocity of light?
Answer:
3 × 10⁸ m/sec

Question 45.
Define wavelength.
Answer:
It is the distance between two successive crests (peaks) or trough in a wave.

Question 46.
Define Wave number.
Answer:
It is the number of waves present per meter and is equal to the reciprocal of wavelength. \(\left(\frac{1}{\lambda}\right)=\bar{v}\)

Question 47.
Define frequency of light.
Answer:
The number of waves which pass through a given point in one second is called frequency of light.

Question 48.
What type of waves does light constitute?
Answer:
Electromagnetic waves.

Question 49.
What is orbital (atom orbital) ?
Answer:
Orbital is a three dimensional region in space around the nucleus in which the probability of finding an electron is maximum.

Question 50.
How many electrons can be accommodated in an orbital?
Answer:
Two electrons.

Question 51.
Write the de Broglie’s equation.
Answer:
\(\lambda=\frac{\mathbf{h}}{\mathbf{m} \mathbf{V}}\) where λ – wavelength of wave, h = Plank’s constant, m = mass of the electron, V – velocity of the electron.

Question 52.
Write Rydberg’s formula.
Answer:

Question 53.
Write the Balmer equation.
Answer:

Question 54.
What are the four prominent lines in Balmer series of hydrogen spectrum ?
Answer:
H_α, H_β, H_γ and H_????

Question 55.
What is the value of Rydberg’s constant?
Answer:
10.97 × 10⁶ m^-1

Question 56.
Give the range of wavelengths of visible light.
Answer:
400 nm to 750 nm.

Question 57.
Give the Rydberg equation where R is Rydberg constant?
Answer:

Question 58.
Name the element whose atom contain six protons in the nucleus.
Answer:
Carbon

Question 59.
Name the different series of hydrogen spectrum.
Answer:
(i) Lyman
(ii) Balmer
(iii) Pachen
(iv) Bracket
(v) Pfund

Question 60.
Name two series of hydrogen spectra which fall in the infra – red region.
Answer:
Brackett series and P-fund series.

Question 61.
Match the following:

Answer:
a-iii, b-iv, c-ii, d-i

1st PUC Chemistry Structure of Atom Two Marks Questions and Answers

Question 1.
Write the difference between isotope & isobars.
Answer:

Question 2
Explain Planks quantum theory.
Answer:
Electromagnetic radiations are emitted, absorbed and propagated discontinuously in the form of small packet of energy called photon. A body emits (absorbs) radiations in the integral multiples of ‘photon’.
Energy associated with each photon is E = hv, h = 6.626 × 10 ^-34 Js

Question 3.
Distinguish between emission spectra & atomic spectra.
Answer:
Emission spectra : When an object is strongly heated, it starts to emit light. When the emitted light is subjected to depression, emission spectrum is obtained)
Atomic spectra : When an atom of an element is strongly heated, it starts emitting light energy in different region. This emitted light when dispersed, a large number of closely spaced lines where formed they are called atomic spectra.

Question 4.
What do you mean by electromagnetic spectra ?
Answer:
The complete range of electromagnetic waves in the increasing order of wavelength (decreasing order of frequency) is known as electromagnetic spectra.

Question 5.
Write the electromagnetic spectra in the increasing order of wave length. (Decreasing order of frequency)
Answer:
Increasing order of wave length > Cosmic rays > Gamma rays < X-rays < U.V. rays < Visible rays < IR rays < Microwave rays < Radio waves (i.e, decreasing order of frequency).

Question 6.
Deduce the de-Broglies matter wave equation.
Answer:
Dual nature of electron – de-Broglie’s matter wave equation
According to Einstein’s mass – energy relation, E = mc² … (1)
Where m and c are the mass and velocity of an electron respectively.
According to quantum theory, Energy E = hv = (he) / λ … (2)
Where h is Plank’s constant = 6.626 × 10^-34 Js, u is the frequency of moving electron.
Comparing(1)and(2)me2 = (hc)/λ >me = h/λ > = λ = h /me … (3)
Equation (3) is called de-Broglie’s equation.
For an electron its velocity is given as V, So the de-Broglie’s equation for an electron is given by,
λ = h / mv (mv = p = momentum of electron)
λ = h / p is called de-Broglie’s matter – wave equation

Question 7.
Explain the wave nature of light.
Answer:
Wave nature of light: According to Newton, light is a stream of particles which are also known as corpuscles of light. This could not account for the phenomena of interference and deflection (as Rutherford’s model) but justified reflection and refraction.

Hygen suggested in his wave theory that the fight travels in the form of waves, later James Maxwell proposed that fight and other radiations are transmitted, these waves are associated with oscillating electric and magnetic field.

Question 8.
Distinguish between particle and wave.
Answer:

Question 9.
Explain Werner Heisenberg’s uncertainty principle (qualitative)
Answer:
It is impossible to determine both the momentum (particle nature) and position (wave nature) of a moving sub atomic particle simultaneously with absolute accuracy.
Mathematically ∆x × Ap = h / 4π where ∆x = uncertainty in position :
∆p = uncertainty in momentum ; h = Plank’s constant = 6.626 × 10^-34 Js.

Question 10.
Mention the Merits of Bohr’s theory.
Answer:

1. Explains the formation of hydrogen spectrums
2. It explains the stability of atom
3. Ryd berg constant is calculated by it.

Question 11.
Write any two limitations of Bohr’s theorem.
Answer:

1. Not explained spectra of atoms containing more than one electrons.
2. Not explained fine spectra (when spectroscope of high resolving power is used, it is found that each line in the ordinary specrum spilts into number of components differing in frequency).
3. Not given explanation regarding Stark effect (spiltting of spectral line in electric field) and Zeeman Effect (splitting of spectral fine in magnetic field).

Question 12.
Write the difference between orbit & orbital.
Answer:

Question 13.
Draw the structure of p-orbitals. (Draw the shape of orbital whose Azimuthal quantum no is 1).
Answer:
p orbital has 3 orientations p_x,p_y &p_z

Question 14.
Draw the structure of d-orbital (Orbital whose Azimuthal quantum no = 2).
Answer:
d- orbital has 5 orientations.

Question 15.
What do you mean by electronic configuration ? With the sequence.
Answer:
Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element.
The sequenced in which the electron occupy the various oribitals is as follows.
Is, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p….

Question 16.
Explain the electronic Configuration of cation Fe+.
Answer:
Example for electronic configuration of cation.

Question 17.
Explain electronic configuration of anion using N.
Answer:
Example for electronic configuration of anion.

Question 18.
State & Explain Paul’s Exclusion Principle.
Answer:
Paul’s Exculsion principle : No two electrons in the same atom can have all the four quantum numbers same.
Illustration : In Is orbital there are two electrons, the set of quantum number of one electron differ with another in spin quantum number.

Question 19.
State and explain Hunds Rule of maximum multiplicity.
Answer:
Hund’s rule or maximum multiplicity : Electron pairing does not take place until orbitals of same energy are singly occupied.

Question 20.
What are quantum number and Name them ?
Answer:
In order to define state energy and location of electron a set of four numbers are used.
These numbers are called quantum numbers.

• Principal quantum number (n)
• Azimuthal quantum number (1)
• Magnetic quantum number (m)
• Spin quantum number (s).

Question 21.
Write all quantum number values for 3s orbital electrons.
Answer:
S – orbital has two electrons.
3s; n = 3; 1 = 0; m = 0; s = + 1/2 for first electron,
n = 3; 1 = 0; m = 0; s = -1/2 for second electron.

Question 22.
An orbital can contain only two electrons. Why ?
Answer:
An orbital can have only two electrons, provided these two electrons have anti parallel spins. This arrangement is represented, as ↓↑

Question 23.
Write the atomic number at an element with outer configuration,
i) 4s1, ii) 3d3 .
Answer:
(i) Electronic configuration for outer configuration 4s¹ = 1s². 2s²,2p⁶, 3s², 3p⁶, 4s¹.
Atomic number =19
(ii) Exact configuration for outer configuration 3d3.
1s², 2s², 2p⁶, 3s²,3p⁶, 4s², 3d³ = 23 Atomic number = 23

Question 24.
Write the electronic configuration of
1. Cl^– 2. Na⁺ ion
Answer:
Electronic configuration
1) Cl^– = 1s², 2s², 2p⁶, 3s², 3p⁶
2) Na+ = 1s², 2s², 2p⁶

1st PUC Chemistry Structure of Atom Three / Four Marks Questions and Answers

Question 1.
Summarize the Bohr’s Model of an atom,
Answer:
Bohr’s Model of an atom, the postulates are

1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell.
2. The shells are having different energy levels denoted as K, L, M, N
3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

Question 2.
Explain the experimental set up and different series of emission spectrum of hydrogen.
Answer:
Hydrogen Spectrum : Hydrogen spectrum is obtained by exciting electrons present in hydrogen gas using a discharge tube under low pressure and high voltage of current. The electrons present in various atoms of Hydrogen absorb energy and jumps to higher energy levels. Later they return back to the lower energy levels by emitting excess of energy in the form of photons. When these electrons are coming to the lower energy state a series of lines were formed called Hydrogen spectrum. Denoted as H_α,H_β,H_γ and H_????H_ε

• Lyman Series : This series is formed when the electron jumps from 2nd, 3rd, 4th, …. Higher energy level to first (K shell) level by emitting excess of energy in UV region.
• Baimer Series ; This series is formed when the electron jumps from 3rd, 4th, 5th,… higher energy level to 2nd (L shell) level by emitting excess of energy in visible region.
• Paschen Series : This series is formed when the electron jumps from 4th, 5th, 6th higher energy level to 3rd (M Shell) level by emitting excess of energy in Infra Red (IR Region).
• Brackett Series : This series is formed when the electron jumps from 5th, 6 th…. higher energy level to 4th (N shell) level by emitting excess of energy in Infra Red (IR) region.
• Pfund Series : This series is formed when the electron jumps from 6th, 7th …. higher energy level to 5th (0 shell) level by emitting excess of energy in Infra Red (IR) region.

Question 3.
What is Wave number, Frequency and Amplitude ? Give its SI Units.
Answer:
Wave Number ( \(\overrightarrow{\mathrm{v}}\) ) : Number of waves per unit length. Units : m^-1
Frequency (v) ; Number of wave peaks that pass a given point in time unit.
Units: Hz
Amplitude (A): Height of the wave above the central line.

1st PUC Chemistry Structure of Atom Four / Five Marks Questions and Answers

Question 1.
Calculate the wave number, wavelength and frequency first line of hydrogen spectrum Or Calculate the maximum wave length of a line in the Lyman series.
Answer:
First line is Lyman Series, where n₁ = 1, n₂ = 2. Maximum wave length corresponds to minimum frequency i.e., n₁ = 1, n₂ = 2.

R = Rydberg constant = 1.097 × 10⁺⁷ m. n₁ = 1 n₂ = 2

Wave length λ = 0.8227 × 107 = 8.227 × 10⁶ m^-1

Question 2.
Calculate the limiting frequency of Balmer series.
Answer:
Limiting frequency for Balmer series possible when n₁ = 2 and n₂ = ∞.

Question 3.
Calculate the wavelength and wave numbers of the first and second lines in the Balmer series of hydrogen spectrum. Given R = 1.096 × 10⁷m^-1
Answer:
For Balmer series n₁ = 2 , for third line n₂ = 3, for fourth line n₂ = 4
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by

Question 4.
Calculate the wavelength and wave number of H_γ and H_???? line (R = 10.97 × 106 m^-1)
Answer:
For Balmer series n₁ = 2 , for third line n₂ = 5, for fourth line n₂ = 6
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by

= 364 × 10^-9m = 364.9nm
Wavelength of H line = 364.9nm

Question 5.
Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum.
Rydberg constant R = 10.97 × 106 m^-1
Answer:

Question 6.
Calculate the wave number of the spectral line when electron jumps from the seond Bohr orbit to the ground state. R = 1.097 × 10⁷ m^-1.
Answer:

Wavelength of the spectral line = 8.2258 × 10⁶m^-1

Question 7.
In a hydrogen atom, an electron jumps from third orbit to the first orbit. Find out the frequency and wavelength of the spectral line. Given R = 1.097 × 10⁷m^-1

Wavelength of the spectral line = 1025.4 A⁰
Frequency of the spectral line = 2.9257 × 10¹⁵ s^-1

Question 8.
Calculate the wavelength of 2^nd line and limiting line of Balmer series. If wave length of first line of Balmer series is 656 nm.
Answer:

For first line m = 2, m = 3 & λ = 656 nm

For secong line n1 = 2,n2 = 4

∴ Wavelength of the limiting line n₁ = 2, n₂ = ∞

Question 9.
Calculate the wavelength of a wave of frequency 1012 Hz, travelling with the speed of light 3 x 10⁸ ms^-1
Answer:

Question 10.
Calculate the frequency of electromagnetic radiation having the wavelength 3 μ. Calculate the wave number corresponding to it. (1μ = 10^-6 m)
Answer:

Question 11.
Calculate the frequency and energy per quantum of a radiation with a wavelength of 200 nm. (c = 3 × 10⁸ ms^-1 and h = 6.625 × 10^-34 Js)
Answer:

= 1.5 × 10¹⁵ Hz = 1.55 × 10 ¹⁵Hz.
E = hv = 6.625 × 10 ^-34 x 1.5 × 10¹⁵= 9.9375 × 10^-19 J

Question 12.
Calculate the number of photon of light with a wavelength of 6000A that provide I joule of energy.
Answer:

= 0.00331 × 10^-16J = 3.31 × 10^-19 J
Number of photons 3.31 × 10^-19 J make 1 photon
1J = \(\frac{1}{3.31 \times 10^{-19}}\) = 3.021 × 10²⁰ photons

Question 13.
A major line in an atomic emission spectrum occurs at 450 nm. Find the energy decrease, as this photon is emitted.
Answer:

Question 14.
Calculate the wave number, wavelength and frequency of the first line in the Baimer series.
Answer:

For Balmer Series
n₁ = 2 and n₂ = 3 for the first time

Question 15.
The red light of neon signs has a wavelength of 693 nm. Find the energy difference (per mole of atoms) between the two energy levels involved.
Answer:

= 2.8684 × 10¹⁵ J
Energy difference per mole of atoms = 2.8684 × 10¹⁵ × 6.022 × 10²³

Question 16.
Calculate the wavelength of an electron moving with a velocity of 2.5 × 10^-7 ms^-1 h = 6.626 × 10^-34 Js ; mass of an electron = 9.11 × 10^-31 kg.
Answer:

Question 17.
Find the mass of an electrically charged particle moving with a velocity of 3 × 106 ms^-1 and having a de Broglie wavelength of 2Å.
Answer:

Question 18.
Calculate the energy of a photon whose wavelength is 3.864 × 10^-7m.
Answer:
\(\mathrm{E}=\mathrm{h} \vartheta=\frac{\mathrm{h} \mathrm{c}}{\lambda}\)
h = 6.626 × 10^-34
c = 3.0 × 10⁸ ms^-1

λ = 3.864 × 10^-7

Question 19.
Calculate the de Broglie wavelength of (a) an electron of mass 9.11 × 10^-31 kg and moving with a velocity of 1.0 × 106 ms^-1 (b) a bullet of mass 25g moving with a velocity of 100ms^-1
Answer:

You can Download Chapter 15 Waves Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 15 Waves

1st PUC Physics Waves Textbook Questions and Answers

Question 1.
A string of mass 2.50 kg Is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse Jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Mass per unit length of the string,
µ = \(\frac{2.50 \mathrm{kg}}{20 \mathrm{m}}\) = 0.125 kg m^-1
Given tension, T = 200 N.
The speed of wave on the string is given by,
v = \(\sqrt{\frac{T}{\mu}}\) =\(\sqrt{\frac{200}{0.125}}\) = 40 ms^-1
∴ Time taken by the disturbance to reach the other end of the string:
t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{20}{40}\) = 0.5 s

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When Is the splash heard at the top given that the speed of sound In air Is 340 m s^-1? (g= 9.8 ms^-2)
Answer:
Time taken to hear the splash at the top =Time taken by the stone to reach the pond + Time taken by the sound to travel to the top from the base of tower,
i e., T = T₁ + T₂ ……..(1)
T₁ : We know that s = u t + 1 / 2 a t²
u = 0, s = 300 m, a = g = 9.8 ms^-2
⇒ t = \(\sqrt{\frac{2 \mathrm{s}}{\mathrm{a}}}\) = \(\sqrt{\frac{2 \times 300}{9.8}}\) = 7.82 s
∴ T₁ =7.82 s
T₁ : s = vt ⇒ t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{300}{340}\) = 0.88 s
∴ T₂ = 0.88 s
∴ T₁+ T₂ = 7.82 + 0.88 = 8.70 s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension In the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s^-1?
Answer:
Mass per unit length of the wire,
µ = \(\frac{2.10 \mathrm{kg}}{12.0 \mathrm{m}}\) = 0.175 kg m^-1
The speed of wave on a string is
given by, v=\(\sqrt{\frac{T}{\mu}}\)
Given v = 343 ms^-1
∴ T = µv² =0.175 × 343²
= 20588.575 = 2.06×10⁴ N.

Question 4.
Use the formula v = \(\sqrt{\frac{rP}{\rho}}\) to explain why the speed of sound in air

1. Is Independent of pressure,
2. Increases with temperature,
3. Increases with humidity.

Answer:
Given v = \(\sqrt{\frac{\mathrm{rP}}{\rho}}\) ……….(1)
According ideal gas law, P = \(\frac{\rho \mathrm{RT}}{\mathrm{M}}\) , where
\(\boldsymbol{\rho}\) is the density, T is the temperature, M is the Molecular mass of the gas;
R – universal gas constant.
Substituting for P in (1) we get
v = \(\sqrt{\frac{\mathbf{r} \mathbf{R} \mathbf{T}}{\mathbf{M}}}\)
This shows that v is

1. Independent of pressure
2. Increases with temperature i.e. v ∝ \(\sqrt{T}\)
3. We know that the molecular mass of water (18) is less than that of N₂ (28) and O₂ (32). therefore as humidity increases, the effective molecular mass of air decreases and hence velocity increases.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + vt, i.e., y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave :
(a) (x – vt )²
(b) log [(x + Vy)/x₀]
(c) 1/(x + vt)
Answer:
The converse is not true. For any function to represent a travelling wave, an obvious requirement is that the function should be finite at all times and finite everywhere.

Function a) and b) do not satisfy this requirement and hence cannot represent a travelling wave. Only function c) satisfies the condition.

Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of

1. The reflected sound,
2.  The transmitted sound? The speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.

Answer:
We know that λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
where,
λ : wavelength
v : velocity
f: frequency

1.  Reflected sound:
λ = \(\frac{340}{1000}\) = 0.34 m
(∵ v The reflected sound travels in air)

2. Transmitted sound:
λ = \(\frac{1486}{1000}\) = 1.486 m
(∵ v The transmitted sound travels in water)
Note: When sound moves from one medium to other, the frequency does not change.

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
Given
f = 4.2 × 10⁶ Hz
v = 1.7 × 10³ ms^-1
we know that, λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
\(=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.05 × 10^-4 m
∴ wavelength of sound in tissue = 4.05 × 10^-4 m

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x+ π/4 ) where x and y are in cm and t in s. The positive direction of x is from left to right.

1. Is this a travelling wave or a stationary wave? If it is traveling, what are the speed and direction of its propagation?
2. What are its amplitude and frequency?
3. What is the initial phase at the origin?
4. What is the least distance between two successive crests in the wave?

Answer:

1. Travelling wave: The wave is travelling from right to left with a speed of \(\frac{36 s^{-1}}{0.018 \mathrm{cm}^{-1}}\) = 2000 cm s^-1 = 20 ms^-1
2. Amplitude: 3 Frequency : \(\frac{36}{2 \pi}\) = 5.73 Hz
3. Initial phase at origin: π/4
4. Distance between two successive crests in the wave = \(\frac{2 \pi}{0.018 \mathrm{cm}^{-1}}\) = 349.06 cm = 3.49 m.

Question 9.
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion In traveling wave differ from one point to another: amplitude, frequency or phase?
Answer:

All the graphs have same amplitude of frequency but different initial phase. All graphs are sinusoidal.

Question 10.
For the travelling harmonic wave y(x,t) = 2 cos 2π (10t – 0.0080 x + 0.35) where x and y are In centimetre and t In seconds. Calculate the phase difference between oscillatory motion of 2 points separated by a distance of
a) 4 m
b) 0.5 m
c) λ/2
d) 3λ/2
Answer:
Given
k = 2m × 0.008 cm^-1
We know that, k = \(\frac{2 \pi}{\lambda}\) ⇒ λ = \(\frac{2 \pi}{(0.008 \times 2 \pi)}\) = 125cm
Now, (phase difference) = \(\frac{2 \pi}{\lambda}\) × (path difference)
∴ ∆Φ = \(\frac{2 \pi}{\lambda}\) ∆x

Question 11.
The transverse displacement of a string (clamped at Its both ends) is given by y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\). cos \((120 \pi \mathrm{t})\) where x and y are in m and t In s The length of the string is 1.5 m and its mass is 3.0 x10^-2 kg.
Answer the following :

1. Does the function represent a travelling wave or a stationary wave?
2. Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
3. Determine the tension in the string.

Answer:
1.  The function represents a stationary wave.

2.

∴ λ = 3 m , f = 60 Hz, v = \(\frac{120 \pi}{2 \pi} \times 3\) = 180 ms^-1
for both waves.
3.  We know that T = µ v² where µ = mass per unit length of the string
\(=\frac{3 \times 10^{-2} \mathrm{kg}}{1.5 \mathrm{m}}\) = 2 × 10^-2 kgm^-1
∴ T = 2 × 10^-2 × 180² = 648 N

Question 12.
(i) For the wave on a string described in Exercise 15.11, do all the points c n the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude?
Explain your answers.
(ii) What Is the amplitude of a point 0.375 m away from one end?
Answer:
Given :
y (x, t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)

i) Amplitude of wave = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)

Frequency = 60 Hz.
∴ From the above equation of wave, we see that all points on the string oscillate with same phase and frequency except the nodes (endpoints). We can also see that amplitude of the wave changes for different ‘x’.

ii) Amplitude of a point at 0.375 m
i.e., x = 0.375 m
Amplitude = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right)\)
= 0.042 m

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:

1. y= 2 cos (3x) sin (10t)
2. y= 2\(\sqrt{x-v t}\)
3. y = 3 sin(5x – 0.5t) + 4cos(5x-0.5t)
4. y = cos x sin t + cos 2x sin 2t

Answer:
1.  y = 2cos (3x) sin (10 t)
Represents a stationary wave

2.  y = 2\(\sqrt{x-v t}\)
Does not represent any wave.

3.  y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
= 5 sin \(\left[5 x-0.5 t-\tan ^{-1}(3 / 4)\right]\) Represents a travelling wave,

4.  y = cos x sin t + cos 2x sin 2t
Represents superposition of 2 stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. The speed of a transverse wave on the string,
2. The tension in the string?

Answer:
Given :
f = 45 Hz mass of wire
= m =3.5×10^-2 kg
linear mass density
= µ = mass per unit length
= 4 × 10^-2 kg m^-1
∴ length of the wire
\(=\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\)
l = 0.875 m

1.  We vibrate in fundamental mode
⇒ λ = 2l

∴ λ =2×0.875 = 1.75 m
∴ Speed of wave
= v =fλ = 45×1.75
= 78.75 ms^-1

2. We know that
T = µV²
∴ Tension, T = 4×10^-2 × 78.75^-2 = 248.06 N

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
For a closed organ pipe of length ‘l’, the frequency of n^th mode of vibration is
f = (2n – 1)\(\frac{v}{4 l}\) ………….(1)
Thus the (n + 1)^th mode of vibration of closed pipe is
f = (2n + 1)\(\frac{v}{4 l}\) ………….(2)
Given that the closed pipe vibrates at 340 Hz for tube length = 25.5 cm and 79.3 cm.
Let l₁ = 25.5 cm l₂ = 79.3 cm
Equating (1) and (2) we get

On solving we get n = 1.
On substituting in (1) we have
\(\frac{(2 \times 1-1) v}{4 \times\left(\frac{25.5}{100}\right)}=340\) ⇒ v = 346.8 ms^-1

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Given: Here, length of rod, l = 100 cm = 1 m
Fundamental frequency of longitudinal vibrations,
v = 2.53 kHz = 2.53 x 10³ Hz.
In fundamental mode, wavelength, λ= 21 = 2×1= 2m.
∴ speed of sound in the steel rod.
υ= vλ. = 2.53 x 10³ x 2 = 5.06 X 10³ ms^-1 = 5.06 km s^-1.

Question 17.
A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s resonantly excited by a 430 Hz source? Will the same source be In resonance with the pipe if both ends are open? (speed of sound In air Is 340 m s^-1).
Answer:
Given :
l = 20 cm = 0.2 m
v = 340 ms^-1
f = 430 Hz
For a pipe closed at one end :
f = \(\frac{(2 n-1) v}{4 l}\) n = 1, 2, 3 ……
⇒ 430 = \(=(2 n-1) \frac{340}{4 \times 0.2}\)
⇒ n = 1
⇒ Resonance occurs only for first/fundamental mode of vibration.
For a pipe open at both ends,
f = \(\frac{\mathrm{nv}}{2 l}\) n = 1, 2, …..
⇒ 430 = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\)
⇒ n = 0.51
Since n<1, resonance does not occur.

Question 18.
Two sitar strings A and B playing the note ‘Gd’ are slightly out of tune and produce beats of frequency 6 Hz. The tension In the string A Is slightly reduced and the beat frequency Is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what Is the frequency of B?
Answer:
Let f₁ and f₂ be the frequencies of strings A and B respectively.
Given:
f₁ = 324 Hz
Beat frequency = f_b = 6 Hz
∴ f₂= f₁ ± f_b =(324 ±6) Hz
We know that T ∝ v² and v ∝ f
(∵ T=μv² and v=fλ.)
Thus f ∝ \(\sqrt{T}\) …………(1)
Decreasing the tension in string A will decrease the frequency f₁.
Since the beat frequency f_b is reduced to 3 Hz upon decreasing the tension in string A, f₂ = 318 Hz.
Note:
If f₂ = 330 Hz, then the beat frequency would have increased when the tension in string A was reduced.

Question 19.
Explain why (or how):

1. In a sound wave, a displacement node Is a pressure antinode and vice versa,
2. bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
3. A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
4. Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate In gases, and
5. The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer:

1. In a sound wave, the displacement node is a point where the amplitude of oscillation is zero. But at a displacement node, the pressure changes are maximum. Hence it is also a pressure antinode. Similarly, at displacement antinode, the amplitude of oscillation is maximum whereas pressure changes are minimum. Hence displacement antinode is also a pressure node.
2. Bats emit high-frequency ultrasonic waves. These waves are reflected back from the obstacles on their path and are sensed by bats.
3. Because they emit different harmonies which can be easily differentiated by human ears.
4. This is because solids have both shear and bulk modulus of elasticity whereas gas has only bulk modulus of elasticity.
5. A sound pulse consists of waves with different wavelengths. In a dispersive medium, these waves travel with different velocities which distorts the shape of pulse.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz In still air.
1.  What Is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1,
(b) recedes from the platform with a speed of 10 m s^-1?

2.  What Is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given, f = 400 Hz v = 340 ms^-1 v_s = 10 ms^-1 : speed of train .
1.
a) train approaches the platform

b) train recedes from the platform

2.  The speed of sound does not change, i.e., it is 340 ms^-1 for both cases.

Question 21.
A train, standing In a station-yard, blows a whistle of frequency 400 Hz In still air. The wind starts blowing in the direction from the yard to the station at a speed of 10 ms^-1. what are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly Identical to the case when the air Is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given f = 400 Hz v = 340 ms^-1 v_m = 1.0 ms^-1 : speed of air
Since there is no relative motion between the source and the observer,
the frequency of sound for the observer = f =400 Hz.
The wind is blowing in the direction from the yard to the station, the effective speed of sound for an observer at the platform = v + v_m = 350 ms^-1
Wavelength of sound : λ= \(\frac{\left(v+v_{m}\right)}{f}\)
= \(\frac{350}{400}\) = 0.875 m
The situation is not identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1 (= v₀). Because here there is a relative motion between the source and the observer.

Question 22.
A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4)

1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s.

Answer:
Given:
y (x, t) = 7.5 sin (0.005 x + 12t + π /4)
1. At x=1 cm and t=1s
y (1, 1)= 7.5 Sin(0.005 +12 + π /4)
= 7.5 sin (12.005 + π /4)
= 1.67 cm
Velocity of oscillation : v = \(\frac{\mathrm{d}(\mathrm{Y}(\mathrm{x}, \mathrm{t})}{\mathrm{dt}}\)
= \(\frac{\mathrm{d}}{\mathrm{dt}}\) (7.5 sin (0.005 x + 12t + π/4) dt
= 7.5 × 12 cos (0.005 x+ 12t + π/4)
At x = 1 cm and t = 1 s
v = 7.5 × 12 cos(0.005 + 12 + π/4)
= 87.75 cm s^-1
We know that velocity of wave propagation = \(\frac{\mathrm{w}}{\mathrm{k}}\)
Here w = 12 s^-1 and k = 0.005 cm^-1
∴ Velocity of wave propagation
= \(\frac{12 s^{-1}}{0.005 \mathrm{cm}^{-1}}\) = 24 ms^-1

∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation.

2. In a wave, all the points which are separated by a distance ±λ,±2λ ……..
from x = 1 cm will have same transverse displacements and velocity. For the given
wave , λ= \(\frac{2 \pi}{0.005}\) = ±12.56 cm, +25.12
m….From x = 1 cm will have the same displacements and velocity as at x = 1 cm, t = 2s, 5s and 11 s.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
1.  Does the pulse have a definite
(i) frequency,
(ii) wavelength,
(iii) speed of propagation?
2.  If the pulse rate is 1 after every 20 s, (that is the whistle Is blown for a split of seconds after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
1. In a nondispersive medium, the wave propagates with definite speed but its wavelength and frequency are not definite
2. No, the frequency of the note is not \(\frac { 1 }{ 20 } \) or 0.05 Hz. 0.005 Hz is only the frequency of repetition of the pip of the whistle.

Question 24.
One end of a long string of linear mass density 8.0 × 10^-3 kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down its transverse displacement y as a function of x and t that describes the wave on the string.
Answer:
Given : Mass per unit length of string = μ = 8×10^-3kgm^-1
f = 256 Hz A = 5 cm : amplitude Tension in the string,
T = 90 kg × 9.8 ms^-2= 882 N
velocity of the wave :
v = \(\sqrt{\frac{T}{\mu}}\)

propagation constant,
k= \(\frac{2 \pi}{\lambda}\) = 4.84 m^-1
Equation of wave y (x, t) = A sin(wt – kx)
∴ y(x, t) = 5 sin(512 π t – 4.84 x)
with x and y in centimetre and t in seconds.

Question 25.
A SONAR system fixed In a submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be1450ms^-1.
Answer:
Given: f = 40 kHz v = 1450ms^-1
Speed of enemy submarine = v₀ = 360 km/ hr
=360 × \(\frac{5}{16}\) = 100 m/s
Apparent frequency of sound waves:
(source at rest and observer moving towards the source)

Now this frequency is reflected by the enemy submarine and is observed by the SONAR
∴ \(\mathrm{f}^{\prime}=\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}} \times \mathrm{f}^{\prime}\)
Note: Here the observer (SONAR) is at rest and the source is moving towards the observer at a speed of 360 km h^-1 = 100 ms^-1 (v_s)
\(\therefore \mathrm{f}^{\prime}=\frac{1450}{1450-100} \times 42.76 \mathrm{k}=45.93 \mathrm{kHz}\)

Question 26.
Earthquakes generate sound waves Inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^-1, and that of the P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 minutes before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur?
Answer:
Let v_p =8 km s^-1 and v_s = 4 k ms^-1
t = time gap between arrival of s and p waves = 4 min = 240 s
t = t_s – t_p = \(\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{s}}}-\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{p}}}\)
where d :
distance of earthquake centre (kms)
∴ 240 = \(\mathrm{d}\left[\frac{1}{4}-\frac{1}{8}\right]\)
∴ d = \(\frac{240}{0.25-0.125}\) = 1920 km

Question 27.
A bat Is flitting about In a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat Is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Given:
f = 40 kHz
velocity of bat = 0.03 velocity of sound i.e., v_s = 0.03 v
Apparent frequency of sound hitting the wall
\(f^{\prime}=\frac{v}{v-v_{s}} \times f=\frac{v}{v-0.02 v} \times f\)
= 1.03 \(f^{\prime}\) = 1.03 × 40 k
= 41.24 k Hz
This freuency f1 is reflected e ft tne wall and is recieved by the bat moving towards the wall
∴ \(f^{\prime \prime}=\frac{\gamma+v_{s}}{v} \times f^{\prime}=\frac{v+0.03 v}{v} \times f^{\prime}\)
= 1.01 \(f^{\prime}\) = 1.03 × 41.24 k
= 42.47 k Hz

1st PUC Physics Waves one mark Questions and Answers

Question 1.
What is meant by a wave?
Answer:
The disturbance set up in a medium is called a wave.

Question 2.
What property of the medium is essential for the propagation of mechanical wave?
Answer:
Elasticity and Inertia.

Question 3.
Which physical quantity does not change when a wave travels from one medium to another?
Answer:
Frequency

Question 4.
What is a progressive wave?
Answer:
A wave, which travels continuously in a medium in the same direction, is called a progressive wave.

Question 5.
If y = 2sin π (40t – 2x) represents a progressive wave. What is its frequency?
Answer:
20 Hz.
Solution:
Given equation is
y = 2 sin π (40t – 2x)
y =2 sin 40π (t – x/20)
Comparing this with y = a sin ω (t -x/v)
ω = 2 πf = 40π
∴ f= 20Hz

Question 6.
Two waves are represented by the equations Y₁ = a sin(ωt- kx) and Y₂ = a cos (ω t – kx). What is the phase difference between them?
Answer:
π/2 radians.
Y₁ = a sin (ω t – kx).
Y₂= a cos (ω t – kx).
= a sin(ω t – kx + π /2).
∴ Phase difference between Y₁ and Y₂ is π/2 rad.

Question 7.
What is meant by ‘Phase of a Particle’ in a wave?
Answer:
The phase of a particle at any instant represents the state of vibration of the particle at that instant.

Question 8.
What is a mechanical wave?
Answer:
A wave which requires a medium for its propagation is called a mechanical wave.

Question 9.
Give an example of a transverse wave.
Answer:
Light waves.

Question 10.
Give an example of a two-dimensional wave.
Answer:
Water waves.

Question 11.
What is a transverse wave?
Answer:
In transverse waves, particles of the medium are vibrating perpendicular to the direction of wave propagation.

Question 12.
What is the angle between the vibration of the particle of medium and direction of propagation of the wave in a transverse wave?
Answer:
90°

Question 13.
How does velocity of sound vary with pressure?
Answer:
The velocity of sound is independent of pressure.

Question 14.
How does velocity of sound vary with temperature?
Answer:
The velocity of sound in a gas is directly proportional to the square root of the absolute temperature.

Question 15.
How does velocity of sound vary with humidity?
Answer:
The velocity of sound increases with increase in humidity.

Question 16.
Why sound travels faster in moist air than in dry air?
Answer:
The density of moist air is less than that of dry air. As the velocity of sound in a gas is inversely proportional to the square root of its density, the velocity of sound in moist air is greater than that in dry air.

Question 17.
A wave has a velocity of 330 ms^-1 at one atmospheric pressure. What will be its velocity at 4 atmospheric pressure?
Answer:
330 ms^-1
[Reason: Velocity of sound in a gas is independent of pressure].

Question 18.
What are the beats?
Answer:
The rise and fall in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is known as beats.

Question 19.
What is beat frequency?
Answer:
The number of beats heard per second is called beat frequency.

Question 20.
Define the beat period.
Answer:
The time interval between two consecutive maxima or minima is called beat period. It is also equal to the reciprocal of beat frequency.

Question 21.
What is the available range of sound frequencies?
Answer:
20 Hz to 20,000 Hz.

Question 22.
By how much does the frequencies of 2 sound sources differ, if they produce 10 beats in 2 seconds.
Answer:
Number of beats per second
\(=\frac{10}{2}=5\)
∴ ∆ f = f₁ ~ f₂ = 5 Hz

Question 23.
Define reverberation.
Answer:
Reverberation is defined as the persistence of audible sound even after the source has ceased to produce the sound.

Question 24.
What is a stationary wave?
Answer:
The wave formed due to the superposition of two identical waves travelling with the same speed in opposite directions is called a stationary wave or standing wave.

Question 25.
What Is fundamental frequency?
Answer:
The vibration of a body with the lowest frequency is called the fundamental frequency.

Question 26.
What are overtones?
Answer:
Frequencies greater than fundamental frequency are called overtones.

Question 27.
What are harmonics?
Answer:
Overtones, which are an integral multiple of the fundamental frequency, are called harmonics.

Question 28.
The length of the vibrating portion of a sonometer wire Is doubled. How does its frequency change?
Answer:
Halved
[Reason: Frequency f ∝ 1//. As / is doubled f is halved].

Question 29.
Give the relation between the fundamental note and overtone in an open pipe.
Answer:
\(f^{\prime}=(n+1) f\)
f – fundamental frequency, n = 1,2, 3….
for 1^st and 2^nd  …. overtones.

Question 30.
What is the distance between a node and the neighbouring antinode of a stationary wave?
Answer:
\(\frac{\lambda}{4}\)

Question 31.
The fundamental frequency produced in a dosed pipe is 500 Hz. What Is the frequency of the first overtone?
Answer:
1500 Hz.
[Solution: Frequency of the first over tone in a closed pipe is \(f^{\prime}\) = 3f = 3 × 500 = 1500 Hz].

Question 32.
Find the distance between node and an adjacent antinode if the wavelength in 4m in a stationary wave.
Answer:
Distance between node and adjacent antinode =\(\frac{\lambda}{4}=\frac{4}{4}\) = 1 m

Question 33.
Explain why is it NOT possible to have interference between the waves produced by 2 sitars?
Answer:
Because the waves produced will not have a constant phase difference.

Question 34.
Which harmonies are present in a closed organ pipe?
Answer:
All the odd harmonies are present.

Question 35.
What will be the resultant amplitude when 2 waves y₁ = a sin ω t are superposed at any point at a particular instant?
Answer:
y = y₁ + y₂ = a sin ω t + a cos ω t
= a (sin ω t + cos ω t)
=\(\sqrt{2} \mathrm{a}(\sin (\omega \mathrm{t}+\pi / 4))\)
∴ Resultant amplitude : \(\sqrt{2}\)a

Question 36.
Write 2 characteristics of a medium which determine the speed of sound waves in the medium.
Answer:
Elasticity and inertia.

Question 37.
State the factors in which the speed of a wave travelling along a stretched Ideal string depends
Answer:
Tension and mass per unit length.

Question 38.
The fundamental frequency of oscillation of a closed pipe is 400 Hs. What will be the fundamental frequency of oscillation of open pipe of the same length?
Answer:
f_e = \(\frac{\mathrm{v}}{4 l}\) = 400 Hz
f_o = \(\frac{\mathrm{v}}{2 l}\) ⇒ f_o = 2 f_e = 800 Hz.

Question 39.
Why is it difficult some times to recognise your friend’s voice on phone?
Answer:
Because of modulation.

Question 40.
Which of the following media can pass longitudinal waves only air, water or Iron?
Answer:
Air

1st PUC Physics Waves two marks Questions and Answers

Question 1.
Explain different types of waves (based on medium)
Answer:
Waves are classified into two types:

1. Mechanical Waves:
Waves, which requires a medium for their propagation are called mechanical waves.
e.g.: Waves on the surface of water, Seismic waves (due to earthquake), Sound waves, Waves on stretched string, waves formed in an air column, shock waves.

2. Non-mechanical Waves:
Waves, which do not require a medium for their propagation are called Non-mechanical Waves.
e.g.: Light waves, heat waves, radio waves, X- rays, ultraviolet rays, Infrared rays, etc.

Question 2.
The equation of a progressive wave is y = 0.2 sin(50t-0.5x). Find the amplitude and the magnitude of the velocity, if ‘x’ and ‘y’ are in metres.
Answer:
Given equation is,
y = 0.2 sin(50t – 0.5x)
= 0.2 sin50(t – x/100)
Comparing with y = a sin ω (t – x/v)
amplitude a = 0.2m,
velocity v = 100m/s

Question 3.
State the principle of superposition. Name the phenomenon produced due to the superposition of waves.
Answer:
When two or more waves superpose, he resultant displacement of particle of the medium is equal to the vector sum of the displacements due to individual waves. Superposition of waves leads to the phenomenon of interference, diffraction, beats, and formation of stationary waves are due to the superposition of waves.

Question 4.
What is a longitudinal wave ? Give an example.
Answer:
If the particles of a medium vibrate along the direction of wave propagation then wave is called longitudinal waves,
e.g: Sound waves in air are longitudinal waves.

Question 5.
The distance between two particles is 0.1m; If the phase difference between these points is π/2 rad calculate the wavelength.
Answer:
∆= 0.1m, Φ = π/2 rad

Question 6.
How does its frequency of a tuning fork change when the prongs are

1. filed
2. waxed.

Answer:

1. When the prongs of a tuning fork are filed its frequency increases.
2. The frequency of a tuning fork decreases when the prongs are waxed.

Question 7.
What is meant by beats? What are its applications ?.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats.
The phenomenon of beats can be used.

1. To determine the unknown frequency of a tuning fork.
2. In tuning musical instruments.

Question 8.
What is the Doppler effect? Give an example.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observer is called the Doppler effect.
E.g.: The apparent frequency of the whistle of a train increases as it. approaches an observer on the platform and decreases when the train passes the observer.

Question 9
What are the uses of the Doppler effect?
Answer:

1. Doppler effect is used in a radar system to detect the speed of automobiles and aeroplanes.
2. It is used to determine the speed submarines (Using SONAR).
3. It is used to determine the speed of stars and planets and other celestial bodies.

Question 10.
When two tuning forks A and X are sounded together produces 6 beats per second. If the frequency of A is 341 Hz. What is the frequency of X?
Answer:
f_A = 341 Hz
f=?
and f_b = 6 beats/s
f_b = f_A ~ f_x
or f = f_A ± f_b = 341 ± 6
= 335 Hz or 347 Hz

Question 11.
What are nodes and antinomies in a stationary wave?
Answer:
The points in a stationary wave where the amplitude of vibration of the particles zero are called nodes The points in a stationary wave where the amplitude of vibration of particles is maximum are called antinodes.

Question 12.
An open pipe and a closed pipe have the same fundamental frequency. Explain how their lengths are related?
Answer:
Fundamental Frequency of an open pipe f₁ = \(\frac { v }{ { 2l }_{ 1 } } \)
Fundamental Frequency of an closed pipe
f₂ = \(\frac{\mathbf{v}}{4 \ell_{2}}\)
Given f₁ = f₂ ∵ 2l₁ = 4l₂
\(\frac{\ell_{1}}{\ell_{2}}=\frac{2}{1}\) ⇒ l₁:l₂ = 2:1

Question 13.
Mention any four characteristics of a stationary wave.
Answer:

1. A stationary wave does not move in any direction.
2. There is no flow of energy.
3. All particles in a loop are in the same phase & they are in opposite phase with respect to the adjacent loop.
4. Amplitude is different for different particles.

Question 14.
The fundamental frequency produced in a closed pipe Is 100Hz. What is the frequencies of first and second overtone?
Answer:
Given, f = 100Hz
For the first overtone f_1,
i.e f₁ = 3f
f₁ = 3 × 100
= 300 Hz
For the second overtone f_2,
i.e f₂ = 5f
= 5 × 100
= 500 Hz.

Question 15.
The equation for the transverse wave on a string is y \(=4 \sin 2 \pi(t / 0.05-x / 50)\) with length expressed Iff centimetre and time In second. Calculate the wave velocity and maximum particle velocity.
Answer:
Given

Particle velocity

Question 16.
Equation of a wave travelling on a string is y = 0.1 sin(3001 – 0.01 x) cm. Here x is in centimetre and t is in seconds. Find

1. wavelength of the wave
2. Time taken by the wave to travel 1 m

Answer:

The wave takes T seconds to travel a distance of λ.
∴ Time taken to travel 1 m is

Question 17.
What is meant by RADAR and SONAR? How are long distances measured using these techniques?
Answer:

1. RADAR – Radio Detecting and Ranging.
2. SONAR – Sound Navigation and Ranging

The waves produced by the devices are sent and are reflected by the bodies and reach them back. If the speed of sound is known and the time taken for to and fro journey, the distance can be estimated.

Question 18.
If the frequency of a tuning fork is 256 Hz and speed of sound in air is 320 ms^-1. Find how far does the sound travel when the fork executes 64 vibrations.
Answer:
We know that
v = f λ f= 256 Hz v = 320 ms^-1
∴ λ = \(\frac{\mathrm{v}}{\mathrm{f}} \) = \(\frac{320}{256} m \)
Also, the distance covered in n vibrations = n λ
∴ Distance covered in 64 vibrations = \(\frac{64 \times 320}{256} \) = 80 m

1st PUC Physics Waves four/five marks Questions and Answers

Question 1.
Distinguish between longitudinal and transverse waves.
Answer:
1. The vibration of particles of the medium is along the direction of wave propagation in longitudinal waves, whereas in transverse waves, the vibration of particles of the medium is perpendicular to the direction of wave propagation.

2. The wave propagates by forming alternate compressions and rarefactions in longitudinal waves, whereas in transverse waves, the wave propagates by forming alternative crests and troughs.

3. Longitudinal waves travel through solids, gases, and liquids, whereas Transverse waves travel through solids and on 1 liquid surfaces.

4. Longitudinal waves cannot be polarised, whereas Transverse waves can be polarised.

5. Distance between two successive compressions or rarefactions is equal to the wavelength of the wave in longitudinal waves, whereas, in transverse waves, the distance between two successive crests or troughs is equal to the wavelength of the wave.

Question 2.
Define the following terms.

1. Wave amplitude
2. Wave period
3. Wave frequency
4. Wavelength
5. Wave velocity

Answer:
1. Wave amplitude (a):
The maximum displacement of a particle of the medium from its mean position is called 1 wave amplitude.

2. Wave period (T):
The time taken by a particle of the medium to complete one vibration or Wave period is the time during which one complete wave is set up in a medium.

3. Wave frequency (f):
The number of vibrations completed by a particle of the medium in one second is called wave frequency or Wave frequency is the number of waves set up in the medium in one second.

4. Wavelength (l):
The distance traveled by the wave in a time equal to its period is called wavelength.

5. Wave velocity (v):
It is the distance traveled by a wave in one second.

Question 3.
What are the characteristics of progressive wave?
Answer:

1. A progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave travels with a certain velocity.
3. There is a flow of energy in the direction of the wave.
4. No particles in the medium are at rest.
5. The amplitude of all the particles is the same.
6. Phase changes continuously from par¬ticle to particle.

Question 4.
State Newton’s formula for the velocity of sound in a gas. What is the Laplace’s correction? Explain.
Answer:
According to Newton, velocity of sound in any medium is given by v = \(\sqrt{\frac{E}{\rho}}\) where E is the modulus of elasticity and p is the density of the medium.
For gases E = B, bulk modulus
∴ v = \(\sqrt{\frac{\mathrm{B}}{\rho}}\) …………(1)
When sound waves travel through a gas alternate compressions and rarefactions are produced. At the compression region pressure increases and volume decreases and at the rarefaction region pressure decreases and volume increases. Newton assumed that these changes take place under isothermal conditions i.e., at a constant temperature.
Under isothermal condition, B = P, pressure of the gas.
∴ In (1) v= \(\sqrt{\frac{P}{\rho}}\) ………..(2)
This is Newton’s formula for velocity of sound in a gas.
For air at NTP, P = 101.3 kPa and
ρ = 1.293 kgm^-3
Substituting the values of P and ρ in . equation (1) we get v = 280m/s. This is much lower than the experimental value of 332 m/s. Thus Newton’s formula is discarded.

Laplace’s correction:
According to Laplace, in a compressed region temperature increases and in a rarefied region it decreases and these changes take place rapidly. Since air is an insulator, there is no conduction of heat. Thus changes are not isothermal but adiabatic.
Under adiabatic condition, B = γ P, where γ is the ratio of specific heats of the gas.
Substituting in equation (1), v = \(\sqrt{\frac{\gamma P}{\rho}}\)
The above equation is called Newton- Laplace’s equation
Substituting the values of P, ρ and γ in the above equation, gives the velocity of sound in air at NTP to be about 331 m/s. This is in close agreement with the experimental value.

Question 5.
Discuss the variation of velocity of sound with,

1. Pressure
2. Temperature
3. Humidity
4. Wind

Answer:
1. Effect of Pressure:
According to Boyle’s law, for the given mass of a gas, pressure (P) is inversely proportional to its volume (V) at constant temperature
\(P \propto \frac{1}{V}\)
or PV = constant if m is the mass and
ρ is the density of a gas then,
\(\mathrm{P}\left(\frac{\mathrm{m}}{\rho}\right)\) = constant
For given mass of gas \(\frac{P}{\rho}\) = constant.
∴ In the equation v = \(\sqrt{\frac{\gamma P}{\rho}}\)
γ and \(P / \rho\) are constants.
Thus velocity of sound is independent of pressure at constant temperature,

2. Effect of Temperature:
From Charle’s law, for the given mass of a gas, the volume V is directly proportional to its absolute temperature T at constant pressure.

At constant pressure, velocity of sound

Hence velocity of sound in a gas is directly proportional to the square root of the absolute temperature,

3. Effect of Humidity:
The presence of water vapour (humidity or moisture) in the air reduces the density of air.
∴ The density of dry air is greater than the moist air.
As the velocity of sound in a gas is inversely proportional to the square root of the density, the velocity of sound in moist air is greater than that in dry air. Thus, as the humidity increases, the velocity of sound also increases.

4. Effect of wind:
The velocity of sound is greater in the direction of the wind and lesser in the opposite direction. Let v be the velocity of sound waves and v_w the velocity of wind if the wind blows in the direction of sound waves, then the resultant velocity of sound is (v + v_w). If the wind blows against the velocity of sound waves, the resultant velocity of sound is (v- v_w).

Question 6.
Explain the theory of beats.
Answer:
Consider two sound waves of the same amplitude ‘a’ and slightly different frequencies f₁ and f₂ travelling in the same direction. The displacement of a particle in a time t due to the two waves is y₁ = a sin ω₁ t, and y₂ = a sin ω₂ t The resultant displacement of the particle due to the superposition of these two waves is,


Is the amplitude of the resultant wave. The intensity of resultant sound is maximum when A is maximum.
A is maximum when cos 2 π \(\left(\frac{f_{1}-f_{2}}{2}\right) t=\pm 1\)

The interval between successive maxima is \(\frac{1}{t_{1}-t_{2}}\),
The number of times intensity of sound becomes maximum per second is f_B = \(\frac{1}{\mathrm{T}_{\mathrm{B}}}\) = f₁ – f₂
Hence the beat frequency is the differ-ence between the frequencies of the two waves.

Question 7.
Derive a general expression for apparent frequency when the source moves towards the observer and observer moving away from the source.
Answer:

Consider a source S emitting sound of frequency f. Let v be the velocity of sound. Let the source move towards the observer with velocity vs and the observer move away from the source with velocity vQ. In one second the source travels a distance SS’ = v_s and wave travels a distance SP = v. In one second source emits f waves such that these waves will be contained in a length S’ P = v – v_s.
The apparent wavelength of these waves is

These waves approach the observer O with a relative velocity (v – v₀)
∴ Number of waves received by the observer in one second or apparent frequency is,

This is the general expression for apparent frequency.

Question 8.
What is Doppler’s effect? Give the expression for the apparent frequency of the note at different cases.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observes is called the Doppler effect.

Case (i) :
When the source moves towards the observer and observer moves away from the source.

Case (ii) :
When the source and ob¬serves move towards each other.

Case (iii) :
When the’source and ob-serves move away from each other

Case (iv):
Source moving away from the observer and the observer moving towards the source

Case (v):
When the source is in motion and the observer is at rest.
a. when the source moves towards the observer

b. when the source moves away from the observer

Case (vi):
when the source is at rest and the observer is in motion a. when the observer moves towards the source

when the observer moves away from the source

where f – real frequency
\(f^{\prime}\) – apparent frequency
v – velocity of sound
v_o – velocity of observer
v_s – velocity of source.

Question 9.
What are the beats? Define beat frequency. Explain how the frequency of a tuning fork is determined using beats.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats. The number of beats heard per second is called the beat frequency and is equal to the difference in the frequency is of the two sound waves. To determine the unknown frequency of a tuning fork

Step 1:
Consider a tuning fork A of known frequency f and another fork B of unknown frequency \(f^{\prime}\). When A and B are sounded together let m beats are heard/sec,
∴\(f^{\prime}\) = f±m

Step 2:
Let one of the prong of B is loaded with a bit of wax. The two forks are again sounded together and let m be the number of beats heard/sec.
If m’ > m, i.e. betas increases after adding wax, then real frequency of B is \(f^{\prime}\) = f – m
If nr’ < m, i.e. betas decreases or remains same after adding wax, then real frequency of B is \(f^{\prime}\) = f + m

Question 10.
Distinguish between stationary wave and progressive wave.
Answer:

1. A stationary wave is formed by the superposition of two equal progressive waves travelling in opposite directions whereas, a progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave does not travel in any direction in stationary waves whereas, in progressive waves, the wave travels with a certain velocity.
3. There is no flow of energy in stationary waves whereas, in progressive waves, there is a flow of energy.
4. Particles at the node are at rest in stationary waves whereas, in progressive waves, no particles in the medium are at rest.
5. In stationary waves, amplitude is different for different particles, whereas in progressive waves, amplitude of all the particles is the same.
6. In stationary waves, all particles in a loop are in the same phase and they are in opposite phase with respect to particles in adjacent loops, whereas in progressive waves, phase changes continuously from particle to particle.

Question 11.
Derive the equation for a stationary wave.
Answer:
The equation of two waves having the same amplitude, wavelength, and speed but propagating in opposite directions is

Where a is the amplitude, λ is the wave-length and v is the velocity of the wave. A stationary wave is formed due to the superposition of these two waves. The resultant displacement of a particle is given by,
y = y₁ + y₂

where A = 2a cos\(\frac{2 \pi}{\lambda} x\) represents the amplitude of the resultant wave.

Question 12.
What is a closed pipe? Show that the overtones ira closed pipe are odd harmonics of the fundamental.
Ans: A pipe which is open at are end and closed at the other end is called a closed pipe.

Consider a closed pipe of length l. Let v be the velocity of sound in air. The air column in a closed pipe vibrates in such. a way that always antinodes formed at the open end and node is formed at the closed end. Let f₁, f_2, and f₃ be the frequencies and l₁, l₂ art l₃ be the wavelengths of 1^st, 2^nd and 3^rd modes of vibration respectively.

For 1st mode (fundamental mode) l = \(\frac{\lambda_{1}}{4}\) or  λ₁=4l
but \(\mathrm{f}_{1}=\frac{\mathrm{V}}{\lambda_{1}} \text { or } \quad \mathrm{f}_{1}=\frac{\mathrm{V}}{4l}\) ……….(1)
For i2nd mode (1sl overtone)
\(l=\frac{3 \lambda_{2}}{4} \quad \text { or } \quad \lambda_{2}=\frac{4 l}{3}\)
but \(\mathrm{f}_{2}=\frac{\mathrm{V}}{\lambda_{2}} \text { or } \quad \mathrm{f}_{1}=\frac{3 \mathrm{V}}{4l}=3 \mathrm{f}_{1}\) ………(2)
from (1)
For 3rd mode (2nd overtone)
\(l=\frac{5 \lambda_{3}}{4} \quad \text { or } \quad \lambda_{3}=\frac{4 l}{5}\)
but \(\mathrm{f}_{3}=\frac{\mathrm{V}}{\lambda_{3}} \text { or } \quad \mathrm{f}_{3}=\frac{5 \mathrm{V}}{4l}=5 \mathrm{f}_{3}\) …………(3)
from (1)
From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an closed pipe, the frequencies of overtones are odd harmonics of the fundamental.

Question 13.
What is an open pipe Show that overtones In an open pipe are harmonics of the fundamental? OR Discuss the modes of vibration of air in an open pipe. OR Show that all harmonics are present in an open pipe.
Answer:
A pipe which is open at both ends is called open pipe.

Consider an open pipe of length l. Let v be the velocity of sound in air. The air column in an open pipe vibrates in such a way that always antinodes are formed at the open ends. Let f₁ f₂ and f₃ be the frequencies and λ₁, λ₂ and λ₃ be the wavelengths of the 1^st, 2^nd and 3^rd modes of vibration respectively. For first mode (fundamental mode).

From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an open pipe, the frequencies of overtones are simple harmonics of the fundamental.

Question 14.
Derive an expression for fundamental frequency In the case of stretched string.
Answer:
In the fundamental mode of vibration of the string, there will be an antinode in between the two nodes a the fixed points.

If l is the length of the string then

Velocity of the-wave along the string is

where T is the tension and m is the mass per unit length (linear density) of the string. Fundamental frequency of vibration of the string is,

Question 15.
Given below are some examples of wave motion. State In each case If the motion is transverse, longitudinal or a combination of both.

1. Motion of a kink In a long coil spring produced by displacing one end of the spring sideways.
2. Wave produced in a cylinder containing water by moving its piston back and forth.
3. Wave produced by a motorboat sailing in water.
4. Ultrasonic waves In air produced by a vibrating crystal.

Answer:

1. Longitudinal wave
2. Transverse wave
3. Combination of both
4. Longitudinal wave

Question 16.
What do you mean by wave motion? Discuss Its four important characteristics.
Answer:
Wave motion is a motion where the energy is transferred without shifting the material particles.
Four characteristics:

1. It is a Simple Harmonic Motion.
2. Energy is transported without material shift.
3. The velocity of waves depends on the medium (only for longitudinal waves).
4. The particles oscillate in SHM.

Question 17.
A simple harmonic wave Is ex-pressed by the equation

y and x are In centimetre and t in seconds. Calculate the following:
i) amplitude
ii) frequency
iii) wavelength
iv) wave velocity
v) phase difference between two particles separated by 17.0 cm
Answer:


= \(\frac{2 \pi}{5}\) rad.

1st PUC Physics Waves numerical problems Questions and Answers

Question 1.
A transverse wave is represented by y = 5 sin (50 πt – πx). Find the wavelength and velocity of the wave.
Solution:
Given equation is,
y = 5 sin (50 πt – πx)
Comparing with the standard transverse wave equation,

∴ Velocity of the wave, u = 50 m/s

Question 2.
A wave travelling at a speed of 200 m/s has a frequency of 1000 Hz. Its amplitude is 2 units. Write down the wave equation.
Solution:
Standard wave equation is y = a sin 2πt \(\left[\frac{t}{T}-\frac{x}{\lambda}\right]\)
Given a = 2 units.
Frequency = 1/T = 1000 Hz
Velocity, u= 200 m/s

Therefore the wave equation is

Question 3.
The distance between two particles on a string is 10 cm. Find the phase difference between these particles if the frequency of the wave is 400 Hz and speed Is 100m/s.
Solution :
If the distance between two points is ∆x, the phase difference is given by
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)
Here, Wavelength \(\lambda=\frac{v}{f}=\frac{100}{400}=0.25 \mathrm{m}\)
path difference \(\Delta x=10 \mathrm{cm}=0.1 \mathrm{m}\)
∴ Phase difference \(\Delta \phi=\frac{2 \pi}{0.25} \times 0.1\)
= 0.8π radians
=144°.

Question 4.
A sinusoidal wave propagating through air has a frequency of 200Hz. If the wave speed is 300m/s, how far apart are two points in the medium with a phase difference of 45° and 60°?
Solution:
v = 300m/s, f = 200Hz, Φ₁=45°, Φ₂=60°
v = f λ
∴ wavelength \(\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{300}{200}=1.5 \mathrm{m}\)
path difference = \(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points

Question 5.
A wave traveling along a string is described by y(x,t) = 0.00327 Sin(72x-2.72t) in which the numerical constants are in SI units (i.e.,\ 0.00327m, 72.1 rad/m and 2.72 rad/s.) Find the amplitude, wavelength, period and speed of the wave.
Solution:
The equation for a wave travelling along the +ve x direction is
y(x, t) = a sin(kx – ωt)-(1)
The given equation is
y(x, t) = 0.00327sin(72.1 x -2.72t) – (2)
comparing equation (1) and equation (2)
1.  Amplitude a = 0.00327 m

2.  Wavelength:
We have k =72.1 rad m^-1
ω = 2.72 rad s^-1

3.  period, T = \(\frac{2 \pi}{\omega}\)

\(=\frac{2 \pi}{2.72} \quad=2.31 \mathrm{sec}\)

4.  Frequency f = \(\frac{1}{T}\)
\(\frac{1}{2.31}=0.4329 \mathrm{Hz}\)

5.  Speed of the wave: from the equation
v = fλ
= 0.4329×0.0872
= 0.0377ms^-1.

Question 6.
A sinusoidal wave propagating through air has a frequency of 150HZ. If the wave speed is 200 ms^-1 how far apart are two points in the medium which have a phase difference of 45° and 150°?
Solution:
f =150Hz, v = 200ms^-1 φ₁ =45° φ₂=150°
v= fλ
wave length \(\lambda=\frac{v}{f}=\frac{200}{150}=1.333 \mathrm{m}\)
path difference =\(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points is

Question 7.
A wave travelling along a string is represented by the equation,
y = 0.08 Sin (5t – 3x) Where x and y are in metre and t is in second.

1. At t = 0.1 sec, find the displacement at x = 0.2m
2. At x s 0.1m, find the displacement at t = 0.4 sec.

Answer:
1.

(∵sin (- θ) = -sin (θ) and sine function is in radian. It is converted into degree by multiplying by 180/3.14)
y =-7.98x 10^-3m.

2. t = 0.4s and x = 0.1m.
y = 0.08 sin (5t – 3x) = 0.08 sin (5 × 0.4 – 3 × 0.1)
y = 0.08 sin (2 – 0.3) = 0.08 sin (1.7)
y = 0.08 sin \(\left(\frac{1.7 \times 180}{3.14}\right)\) = 0.08 sin (97.45°)
y = 0.08 cos (7.45°) = 0.07932 m
(∵sin (90 + θ) = cos (θ)).

Question 8.
A train moving at a speed of 72kmph towards a situation sounding a whistle of frequency 600 Hz. What are the apparent frequency of the whistle as heard by a man on the platform when the train

1. approaches him?
2. recedes from him? (speed of sound In air Is 340 ms^-1).

Answer:
V_s= 72m/hr = 72x 1000m/3600s,
= 20m/s, V = 340m/s, f = 600Hz

1. Apparent frequency when train approaches the observer
\(f^{\prime}=\left(\frac{V}{V-V_{s}}\right) f=\frac{340 \times 600}{340-20}=637.5 \mathrm{Hz}\)

2. Apparent frequency when train recedes from observer
\(\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{f}=\frac{340 \times 600}{340+20}=566.7 \mathrm{Hz}\)

Question 9.
Find the ratio of velocity of sound in oxygen and velocity of sound In hydrogen at S.T.P. Given the molecular weight of Oxygen is 32 and that of Hydrogen is 2.
Solution :
Let v₀ and v_H be the velocities of sound in oxygen and hydrogen respectively
Then

Where Po and pH are the densities of oxygen and Hydrogen respectively. X is the ratio of specific heats which is same for hydrogen and oxygen. But density is directly proportional to the molecular weight.

Question 10.
At what temperature will the velocity of sound in air reduces to half of its value at 0° C?
Solution :

Question 11.
Two tuning forks X and Y sounded together produce 10 beats per second. When Y is slightly loaded with wax, the number of beats reduces to 6 per second. If the frequency of X is 480 Hz, find that of Y.
Solution :
Frequency of x = 480 Hz
No. of beats with y = 10
∴ Frequency of y = 480+10 or 480-10
i.e., 490 or 470
On loading y, number of beats = 6
Frequency of y after loading = 486 or 474
∴ The frequency of y before loading can not be 470, because if it were 470 before loading, it must be less than 470 after loading.
∴ Actual frequency of y = 490 Hz.

Question 12.
Two tuning forks A and B gives 4 beats per second. The frequency of A is 510 Hz. When B Is filed 4 beats per second are again produced. Find the frequency of B before and after filing.
Solution:
Frequency of A =510 Hz.
Beats per second = 4
Therefore the frequency of B before filing,
= 510 + 4 or 510 – 4
= 514 or 506
After filing B, 4 beats are produced again the frequency of B after filing
=510 + 4 or 510 – 4
= 514 or 506
The frequency of B before filing can not be 514 Hz because if it is 514 before filing, after filing its frequency must be more than 514.
Therefore the frequency of B before filing must be 506 Hz and after filing it is 514 Hz.

Question 13.
Two sound waves of wavelength 1.34 m and 1.36 m produces 4 beats per second. Calculate the velocity of sound in the medium.
Solution :
Let v be the velocity of sound in the given medium. Then Frequency of the first wave
\(f_{1} \quad=\frac{v}{\lambda_{1}}=\frac{v}{1.34}\)
Frequency of the second wave

Velocity of sound v = \(\frac{4 \times 1.34 \times 1.36}{0.02}\)
= 364.5 ms^-1

Question 14.
A set of 65 tuning forks are arranged In the Increasing order of frequencies such that each gives 3 beats per second with the previous one. Find the frequency of the first tuning fork If the frequency of the last tuning fork Is one Octave above the first one.
Solution :
Let N be the frequency of the first tuning fork. As each tuning fork is giving 3 beats with the preceding one, and they are arranged in the ascending order of frequencies,
Frequency of the second tuning fork = N + 3
Frequency of the third T.F. = N + 6
= N + 3 × 2
Frequency of the fourth T.F = N+9
= N + 3 × 3
Similarly Frequency of the 65th T.F. = N + (65-1 ) × 3
= N + 64 × 3 = N+192 ……………….(1)
But the frequency of the last tuning fork (65th) is one octave above that of the first one.
∴ Frequency of the 65th T.F.= 2N ……………(2)
From (1) and (2) 2N = N + 192
N = 192
∴ Frequency of the first tuning fork = 192 Hz.

Question 15.
A source of ultrasonic wave is emitting ultrasonic waves of frequency 30 kHz. It is placed in a moving car. With what velocity is the car moving If the frequency appears to be 20 kHz to a stationary observer? Velocity of sound in car 340 ms^-1.
Solution:
Here the listener is at rest and the source is moving. As the apparent frequency is lesser than the actual frequency, the source is moving away from the listener.
To find the velocity with the source is moving:
The apparent frequency is given by,
\(\mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \quad \therefore \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\)
Here, \(f^{\prime}\) is the apparent frequency = 20 kHz
f is the actual frequency = 30 kHz
v is the velocity of sound = 340 ms^-1
On substituting the values, \(\frac{20}{30}=\frac{340}{340+v_{s}}\)
cross multiplying, 2(340 + v_S) = 3 × 340 2 × 340 + 2v_s = 3 × 340
2 . v_s = 3 × 340 – (2 × 340) or 2 . v_s = 340
v_s =\(\frac{340}{2}\) = 170ms^-1
∴ The car is moving with a velocity of 170ms^-1 away from the listener.

Question 16.
An engine moving with a speed of 25 ms^-1 sends out a whistle at a frequency of 280 Hz. Find the apparent frequency of the whistle for a stationary observer

1. when the engine Is approaching him and
2. when it is moving away from him. The velocity of sound is 330 ms^-1.

Solution :

1. When the source is moving towards the observer,
apparent frequency \mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}
Here source velocity v_s = 25 ms^-1,
velocity of sound v = 330 ms<sup.-1
frequency f = 280 Hz.
\(\therefore \mathrm{f}^{\prime}=\left(\frac{330}{330-25}\right) \times 280=302.95 \mathrm{Hz}\)

2. When the source is moving away from the observer, apparent frequency is given by,

Question 17.
The apparent frequency of the whistle of an engine changes in the ratio 6:5 as the engine passes a stationary observer. If the velocity of sound is 330ms^-1 what is the velocity of the engine?
Solution:
Let f be frequency of the sound emitted by the engine and vs be its velocity. The apparent frequency f of the whistle as the engine is approaching the observer is, \(\mathrm{f}^{\prime}=f \frac{\mathrm{V}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\)
But v = 330 ms^-1
\(\therefore \mathrm{f}^{\prime}=\mathrm{f} \frac{330}{330-\mathrm{v}_{\mathrm{s}}}\)   ………(1)
Let f” be the apparent frequency of the sound heard when the engine is moving away from the observer. Then
But v = 330 ms^-1

Cross multiplying
6 × (330 – v_s) = 5 × (330 + v_s)
330 × 6 – 6V_s = 5 × 330 + 5 v_s
330 × 6 – 330 × 5 = 5v_s + 6v_s
330 (6 – 5) = 11v_s
330 × 1 = 11 v_s
∴ v_s \(=\frac{330}{11}\) = 30 ms^-1

Question 18.
Two cars approach each other with a common speed of 20ms1. The first car sounds a horn and a passenger in the other car estimates it to be 700 Hz. The speed of sound is 332 ms^-1.

1. Calculate the actual frequency of the horn
2. When the cars move away from each other, what Is the estimated frequency by the same passenger?

Answer:
1. To find true frequency f:

2.

Question 19.
An observer standing by the side of a highway estimates a drop of 20% in the pitch of the horn of a car as it crosses him. If the velocity of sound is 330 m/s, calculate the speed of the car.
Answer:
Given :
v = 330 m/s
Let f₁ and f₂ be the apparent frequencies heard ty the observer, before and after the source crossing respectively.


Solving this we get vs = 36.67 ms^-1

Question 20.
A person standing in front of a mountain at a certain distance beats a drum at regular intervals. The drum¬ming rate is gradually increased, and he finds that the echo is not heard distinctly when the rate becomes 50 per minute. He then moves nearer to the mountain by 100 meters, and finds that the echo is again not heard when the drumming rates become 60 per minute. Calculate:

1. The distance between the mountain and the initial position of the man,
2. Velocity of sound.

Answer:
Let ‘s’ be the distance between the man and the mountain and let ‘v’ be the velocity of sound.
Given :
Drumming rate = 50 per minute.
∴ Interval between successive beats
\(=\frac{60}{50}=1.2 / \mathrm{sec}\)
Time taken by the echo \(=\frac{2 \mathrm{s}}{\mathrm{v}}\)
Given that when the drumming rate is 50 per minute, echo is not heard by the man ⇒ beats overlap in time frame

Similarly when the person moves 100 m
towards the mountain, with drumming rate = 60 per minute

Substituting this in (1) we get.
v = 1000 m/s

Question 21.
An open pipe Is 30 cm long. Find the fundamental frequency of vibration. Which harmonic is excited by a tuning fork of frequency 22 kHz? velocity of sound is 340 ms^-1.
Solution:
The fundamental frequency of an open pipe is

The frequency of the n^th mode of vibration is given by

Thus the 2.2 kHz source will produce 4^th harmonic.

Question 22.
Two open pipes when sounded together produce 10 beats. If the lengths of the pipes are In the ratio of 4:5, calculate their frequencies.
Solution :
Let f₁ and f₂ be the fundamental frequencies of the two pipes and L₁ be their lengths.
Then,

Then,

Question 23.
Two tuning forks A and B gives 6 beats per second. A. is in resonance with a closed pipe of length 15 cm. and B Is In resonance with an open pipe of length 30.5 cm. Calculate the frequencies of A and B.
Solution:
Let f₁ and f2 be the frequencies of the tuning forks A and B respectively.
Then, f₁ – f₂ = 6 ………….(1)
But tuning fork A is in resonance with a closed pipe of length 15 cm
\(\therefore f_{1}=\frac{V}{4 L_{1}}=\frac{v}{4 \times 0.15}=\frac{v}{0.6}\)
Similarly tuning fork B is in resonance with an open pipe of length 30.5 cm

Frequency of the tuning fork A is
\(\mathrm{f}_{1}=\frac{\mathrm{v}}{0.6}=\frac{219.6}{0.6}=336 \mathrm{Hz}\)
Frequency of the tuning fork B is
\(\mathrm{f}_{2}=\frac{v}{0.61}=\frac{219.6}{0.61}=360 \mathrm{Hz}\)

Question 24.
A closed pipe resonates In its fundamental mode of frequency 500 Hz In air. What will be the fundamental frequency if air is replaced by hydrogen at the same temperature? Density of air = 1.20 kg/m³ and density of hydrogen = 0.089 kg/ m³.
Solution:
Let v_a be the velocity of sound in air and v_H be the velocity of sound in hydrogen. Then fundamental frequency of the pipe filled with air is,

If f is the fundamental frequency of the closed pipe, filled with hydrogen then,

Where γ is the ratio of specific heats, P is the pressure and Pa is the density of air.
Similarly, \(v_{H}=\sqrt{\frac{y P}{\rho_{H}}}\)
\(\therefore \quad \frac{v_{\mathrm{a}}}{v_{\mathrm{H}}}=\sqrt{\frac{\rho_{\mathrm{H}}}{\rho_{\mathrm{a}}}}=\sqrt{\frac{0.089}{1.20}}\) =0.2734
On substituting in equation (3),
\(\frac{500}{f}=0.2734\)
\(f=\frac{500}{0.2734}\)
=1836 Hz.

Question 25.
A string vibrates with a frequency of 200 Hz. When its length Is doubled and Its tension is altered it begins to vibrate with a frequency of 300 Hz. What Is the ratio of new tension to the original tension?
Solution:
\(f=\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
Let L be the length of the string and m be the mass per unit length and T₁ be the original tension. Then,
\(200=\frac{1}{2 L} \sqrt{\frac{T_{1}}{m}}\) …………..(1)
In the Second case length = 2L, let T₂ be the tension

Question 26.
Two tuning forks A and B when vibrated together gives 6 beats per second. The tuning fork A is in unison with an air column in a closed pipe 0.15m long vibrating In fundamental mode and tuning fork B is in unison with an air column In an open pipe 0.61 m long vibrating In first overtone. Calculate the frequencies of the tuning forks.
Solution:
Let f and f be the frequencies of tuning forks A and B respectively.

\(\frac{\omega 1}{m} f_{m}-f_{c}=6 \quad \Rightarrow \frac{f_{B}}{f(1)}=6\)
f_B = 360 Hz
∴ f_A = 6+f_B = 366 Hz.

Question 27.
Two tuning forks, when sounded together, produce 5 beats per second. A sonometer wire of length 0.24 m is in unison with one of them. If the length of the wire is increased by 0.01 m, ft is in unison with the other fork. Find the frequencies of the forks.
Solution:
Let f₁ and f₂ be the frequency of the two turning forks.
given f₁ ~ f₂ =5 ……….(1)
l₁ = 0.24m, l₂ = 0.25m.
Frequency of vibration of the wire is given by

Question 28.
A closed organ pipe of length 0.42 m and an open organ pipe, both contain air at 35° C. The frequency of the first overtone of the closed pipe is equal to the fundamental frequency of open pipe. Calculate the length of open pipe and the velocity of sound in air at 0° C. Given that closed pipe is in unison in its fundamental mode with a tuning fork of frequency 210 Hz.
Solution:
Let l₁ be the length of the closed pipe and l₂ be the length of the open pipe. l₁ = 0.42m
Let v_t be the velocity of sound in air at t° c, t = 35° c
Let f be the fundamental frequency of the closed pipe
\(t=v_{1} / 4 / 1\)
Let f be the fundamental frequency of the open pipe
\(f^{\prime}=v_{1} / 2 l_{2}\)
Given that frequency of first overtone of the closed pipe is equal to the fundamental frequency of the open pipe

Also given that closed pipe is in unison with a tuning fork of frequency 210Hz i.e.,
f = 210 Hz
∴ In (1) 210 \(=\frac{v_{t}}{4 \times 0.42}\)
v_t=840×0.42 =352 .8m /s
Velocity of sound in air at 0°C is
\(v_{0}=v_{t} \sqrt{\frac{273}{273+t}}\)
\(=3528 \sqrt{\frac{273}{273+35}}=3322 \mathrm{m} / \mathrm{s}\)

Question 29.
A stretched wire emits a note of fundamental frequency 35 Hz. When the tension Is increased by 0.5 kg. wt., the frequency of the fundamental rises to 40 Hz. Find the initial tension and the length of the wire. (Mass per unit length of the wire = 1.33 × 10^-3 kg/m).
Solution:
Fundamental frequency of vibration in a stretched string is

Question 30.
One end of a horizontal wire is fixed and the other passes over a smooth frictionless pulley and has a heavy body attached to it. The fundamental frequency is 400 Hz. When the body is totally immersed in water, the frequency drops to 350 Hz. Find the density of the body. \(\left[\rho_{\omega}=19 / \mathrm{cm}^{3}\right]\).
Answer:
Given :
f₁= 400 Hz. f₂ = 350 Hz
\(\rho_{\omega}=^{1} 9 / \mathrm{cm}^{3}\)
We know that, for a string under tension T, the frequency of oscillation
\(\mathrm{f}=\frac{\mathrm{n}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
l: length of wire, \(\mu\) : mass per unit length
∴ T₁ = mg   T₂ = mg – B
Now m: \(\mathrm{v} \rho\)   v: volume      ρ: mass density
∴ mg = \(\mathrm{v} \rho\)g     B = \(\mathrm{v} \rho\)_ωg
g = 9.8 m s^-2
Given:

Question 31.
A whistle, emitting a sound of frequency 300 Hz is tied to a string of 2 m length and is rotated with an angular velocity of 15 rad / second in the horizontal phase. Find the range of frequencies heard by the observer stationed at a large distance from the whistle.
Answer:
Given:
radius r = 2 m ;
w = 15 rad s^-1;  v = 330 ms^-1
We know that v_s = r w = 30 m s^-1

The observer will receive maximum frequency when the source is approaching (A) and minimum when its receding (B).

∴ Frequency range = 275 Hz – 330 Hz.

Question 32.
The wavelengths of 2 notes are 7/165^m and 7/167^m. Each note produces 5 beats per second with a 3^rd note of a fixed frequency. Calculate the velocity of sound in air.
Answer:
Given

Question 33.
An open pipe is suddenly closed at one end with the result that the frequency of 3^rd harmonic of closed pipe is found to be higher by 50 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of open pipe?
Answer:
Let f0 be the fundamental frequency of the pipe of length l. Then,
\(\mathrm{f}_{0}=\frac{\mathrm{v}}{2 l}\) ………..(1)
Let f_e be the 3rd harmonic of closed pipe then,

From (3) and (4) we get f₀ = 100 Hz

Question 34.
Calculate the number of beats heard per second if there are 3 sources of frequencies (n -1), n and (n +1) Hz of equal intensity sounded together.
Answer:
Let us assume each disturbance has an amplitude ‘A’ then the resultant displacement is given by,
y = A sin 2π(n – 1)t + A sin2πnt+ A sin2π(n + 1)t
i.e. y= 2A sin 2πnt cos2πt + A sin2πnt
y= A(1 + 2cos2πt) sin2πnt
∴ Resultant amplitude: A(1 + 2 cos2πt)
Amplitude is maximum when cos2πt = 1
i.e., when 2πt = 2πk k = 0,1,2……….
i.e., when t = 0, 1, 2, 3……….
∴ Time difference between successive maxima = 1 s
Similarly, amplitude is ‘0’ when cos2πt = -1/2
i.e., when cos2πt = 2πk + 2π/3
k = 0, 1, 2………….
i.e. when t = k + 1/3
i.e., when t = 1/3, 4/3, 7/3 …….
Again the time difference between successive minima = 1 s
∴ The frequency of beats is also 1 Hz.
Thus, one beat is heard per second.

Question 35.
Two sound waves originating from the same source, travel along different paths in air and then meet at a point. If the source vibrates at a frequency of 2 kHz and one path is 41.5 cm longer than the other, what will be the nature of interference ? The speed of sound in air is 332 m/s
Answer:
We know that v = fλ
\(\therefore \lambda=\frac{v}{f}\)
Given v = 332 ms^-1
f = 2 k Hz
⇒ \(\lambda=\frac{332}{2 \times 10^{3}}=0.166 \mathrm{m}\)
We know that phase difference (∆Φ)) and path difference (∆x) are related by
\(\Delta \phi=\frac{2 \pi}{\lambda} \quad \Delta x \Rightarrow \Delta \phi=\frac{2 \pi}{0.166} \times 0.415\)
∴ \(\Delta \phi=5 \pi\)
Since phase difference is an odd multiple ot π, the interference is destructive.

Question 36.
A metallic rod of length 2 m is rigidly clamped at its midpoint. Longitudinal stationary waves are set up in the rod in such a way that there are 2 nodes on either side of the midpoint. The amplitude of an antinode is 4 × 10^-6 m. Write an equation of motion of the constituent waves in the rod. (Young’s modulus = 2×10¹¹Nm^-2 and density = 8000 kg m^-3)
Answer:
General equation of a standing wave is y = 2A sink x cos ωt where
\(\mathrm{k}=\frac{2 \pi}{\pi}\) and \(\omega=2 \pi \mathrm{f}\)
Which is obtained by adding 2 identical progressive waves travelling in opposite directions
i.e., y = y₁ + y₂ where
y₁ = A sin (kx – ωt)
y₂ = A sin (kx + ωt)
Let l be the length of the rod: l = 2 m
From the figure below, we see that
\(l=\frac{5 \lambda}{2}\)

We know that velocity of longitudinal wave is given by

∴ Equation of standing wave :
y = 2(4 × 10^-6) sin (2.5 πx) cos (12.5 × 10³π) t
Equation of constituent waves
y₁ = (4 ×10^-6) sin (2.5πx – 12.5 × 10³ πt)
y₂= (4 × 10^-6) sin (2.57πx + 12.5 × 10³ πt)