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KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.5.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational.
Answer:
i) \(2-\sqrt{5}\)
ii) \((3+\sqrt{23})-\sqrt{23}\)
iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
iv) \(\frac{1}{\sqrt{2}}\)
v) 2π
Answer:
i) \(2-\sqrt{5}\) = 2 – 2.2360679……….. = -0.2360679
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

ii) \((3+\sqrt{23})-\sqrt{23}\)
= \(3+\sqrt{23}-\sqrt{23}\)
= 3
⇒ \(\frac{3}{1}\). This can be written in the form of \(\frac{p}{q}\).
∴ This is a rationl number.

iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2 \sqrt{7}}{\sqrt{7}}=2 \Rightarrow \frac{2}{1}\)
This can be written in the form of \(\frac{p}{q}\)
∴ This is a rational number.

iv) \(\frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.4142}{2}\)
= 0.707106…………
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

v) 2π
= 2 × 3.1415…..
= 6.2830…….
This is a non-terminating, non-recurring decimal.
∴ This is an irrational number.

Question 2.
Simplify each of the following expressions:
i) \((3+\sqrt{3})(2+\sqrt{2})\)
ii) \((3+\sqrt{3})(3-\sqrt{3})\)
iii) \((\sqrt{5}+\sqrt{2})^{2}\)
iv) \((\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Answer:

Question 3.
Recall, 2π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, \(\pi=\frac{c}{d}\). This seems to contradict the fact that n is irrational. How will you resolve this contradiction?
Answer:
There is no contradiction. When we measure a length with a scale or any other device we get the quotient.
Therefore we cannot judge whether c is a rational number of d is an irrational number.
∴ Value of \(\frac{c}{d}\) is irrational number.
∴ The value of π is also an irrational number.

Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Answer:
Construction: Mark the distance 9.5 units from a fixed point O such that OB = 9.3 units. Mark midpoint D of OC. Draw a semicircle with centre D. Draw a line perpendicular to OC passing through E and intersecting the semicircle at E. Draw an arc BE which intersect at F. Now, BF = \(\sqrt{9.3}\).

Question 5.
Rationalise the denominators of the following :

Answer:
i) \(\frac{1}{\sqrt{7}}\)
Denominator’s factor is \(\sqrt{7}\) Mulitplying Numerator and denominator by \(\sqrt{7}\).

ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Denominator’s factor is \(\sqrt{7}+\sqrt{6}\)
Multiplying numerator and denominator by \(\sqrt{7}+\sqrt{6}\),

iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
Denominator’s factor is \(\sqrt{5}-\sqrt{2}\)
Multiplying numerator and denominator by \(\sqrt{5}-\sqrt{2}\).

iv) \(\frac{1}{\sqrt{7}-2}\)
Denominator’s factor is \(\sqrt{7}+2\)
Multiplying numerator and denominator by \(\sqrt{7}+2\).

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.5, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.4.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.4

Question 11.
Visualize 3.765 on the number line, using successive magnification.
Answer:
Visualization of 3.765 on the number line:

We know that 3.765 lies between 3 and 4. So, divide the portion between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Then, 3.765 lies between 3.7 and 3.8 Fig. Now, we imagine dividing this again into 10 equal parts. The first mark will represent 3.71, the next 3.72, and so on. To see this clearly, we magnify this as shown in Fig. Then, 3.765 lies between 3.76 and 3.77 Fig. So, let us focus on this portion of the number line Fig and imagine to divide it again into 10 equal pans Fig. Here, we can visualize the 3.761 is the first mark, 3.762 is the second mark, and so on.

Question 2.
Visualise \(4 . \overline{26}\) on the number line, up to 4 decimal places.
Answer:
Visulaisation of \(4 . \overline{26}\) upto 4 decimal places on number line :

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.3.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has :

Answer:
(i)

Terminating decimal expansion.
(ii)

Non-terminating decimal expansion.
\(\begin{aligned} \therefore \quad \frac{1}{11} &=0.090909 \ldots \\ &=0 . \overline{09} \end{aligned}\)
(iii)

∴ Terminating decimal expansion.
(iv)

\(0 . \overline{23069}\) the bar above the digits indicates the block of digits that repeats.
∴ This is Non-terminating, repeating decimal.
(v)

Here, \(0 . \overline{18}\) is block of digits that repeats.
∴ This is non-terminating, repeating decimal.
(vi)

\(\frac{329}{400}=0.8225\) This is terminating decimal because remainder is zero.

Question 2.
You know that\(\frac{1}{7}=0 . \overline{142857}\). Can you predict what the decimal expansions of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) doing the long division ? If so, how? (Hint: Study the remainders while finding the value of \(\frac{1}{7}\) carefully.).
Answer:


Here the first digit in decimal continued repeated numbers afterward.

Question 3.
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0. (p, q ϵ Z, q ≠ 0).
(i) \(0 . \overline{6}\)
(ii) \(0 . \overline{47}\)
(iii) \(0 . \overline{001}\)
Answer:
(i) \(0 . \overline{6}\) = 0.6666……….
Let x = 0.6666………….
∴ x = 0.6666………..
Multiplying both sides by 10,
10x = 6.666…..
10x = 6 + 0.666
10x = 6 + x
10x – x = 6
9x = 6
\(x=\frac{6}{9}=\text { form of } \frac{\mathrm{p}}{\mathrm{q}}\)
∴ \(0 . \overline{6}=\frac{2}{3}\)

(ii) \(0 . \overline{47}\) = 0.4777…..
Let x = \( 0 . \overline{47}\) then x = 0.4777……..
Here digit 4 is not repeating but 7 is repeated.
Let x = 0.4777
Multiplying both sides by 10.
10x = 4.3 + x
10x -x = 4.3
9x = 4.3
Multiplying both sides by 10.
90x = 43
∴ \(0.4 \overline{7}=\frac{43}{90}\)

(iii) \(0 . \overline{001}\) x = 0.001001……….
Multiplying both sides by 1000
1000x = 1.001001
1000x = 1 + x
1000x – x = 1
999x = 1
\(x=\frac{1}{999}\)
∴ \(0 . \overline{001}=\frac{1}{999}\)

Question 4.
Express 0.99999…….. in the form \(\frac{p}{q}\) , Are you surprised by your answer ? With your teacher and classmates discuss why the answer makes sense.
Answer:
Express 0.99999 in the form of \(\frac{p}{q}\),
Let x = 0.99999 …
Multiplying by 10.
10x = 9.9999……
10x = 9 + 0.9999…….
10x = 9 + x
10x – x = 9
9x = 9
\(x=\frac{9}{9}, x=1=\frac{1}{1}\)
∴ (p = 1 & q = 1)
0.99999 …… goes on forever, so there is no gap between 1 and 0.99999 ………… hence they are equal.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \(\frac{1}{17}\) ? Perform the division to check your answer:
Answer:
Expansion of \(\frac{1}{17}\) in decimal:

\(\therefore \quad \frac{1}{17}=0 . \overline{0588235294117647}\)
Here repeated numbers are 16.

Question 6.
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (q≠0), where p
and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer:
Rational numbers in the form \(\frac{p}{q}\) (q ≠ 0) where P and q are integers with no common factors other than 2 or 5 or both.
OR
Prime factors of q have powers of 2 and prime factors of q have powers of 5 or both.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
Three numbers whose decimal expansions are non-terminating, non-recurring are
i) 0.123123312333
ii) 0.20200200020000
iii) 0.56566566656666

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) & \(\frac{9}{11}\) .
Answer:
\(\frac{5}{7}=0 . \overline{714285} \quad \frac{9}{11}=0 . \overline{81}\)
Three irrational numbers between \(0 . \overline{714285}\) and \(0 . \overline{81}\) are :
i) 0.72720720072000………….
ii) 0.7357355735555……………..
iii) 0.760760076000………………

Question 9.
Classify the following numbers as rational or irrational :
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001…
Answer:
i) \(\sqrt{23}\) = 4.7958………. This is not terminating or non-recurring decimal.
∴ \(\sqrt{23}\) is an irrational number.
ii) \(\sqrt{23}\) = 15 this is a rational number.
iii) 0.3796 This is rational number, because trminating decimal has expansion.
iv) 0.478478… This is rational number, because decimal expansion is recurring.
v) 0.101001000100001… This is an irrational number because termianting or recruring decimal has no expansion.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.2.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.2

Question 1.
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Answer:
True. Because set of real numbers contain both rational and irrational number.

(ii) Every point on the number line is of the form \(\sqrt{\mathrm{m}}\). where’m’ is a natural number.
Answer:
False. Value of \(\sqrt{\mathrm{m}}\) is not netagive number.

(iii) Every real number is an irrational number.
Answer:
False. Because set of real numbers contain both rational and irrational numbers. But 2 is a rational number but not irrational number.

Question 2.
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rationed number.
Answer:
Square root of all positive integers is not an irrational number.
E.g. \(\sqrt{\mathrm{4}}\) = 2 Rational number.
\(\sqrt{\mathrm{9}}\) = 3 Rational number.

Question 3.
Show how \(\sqrt{\mathrm{5}}\) can be represented on the number line.
Answer:
\(\sqrt{\mathrm{5}}\) can be represented on number line:

In the Right angled ∆OAB ∠OAB = 90°.
OA = 1 cm, AB = 2 cm., then
As per Pythagoras theorem,
OB² =OA² + AB²
= (1)² + (2)²
= 1 + 4
OB² = 5
∴ OB = \(\sqrt{\mathrm{5}}\)
If we draw semicircles with radius OB with ‘O’ as centre, value of \(\sqrt{\mathrm{5}}\) on number line
\(\sqrt{\mathrm{5}}\) = OM = +2.3
and \(\sqrt{\mathrm{5}}\) = ON = -2.3 (accurately).

Question 4.
Classroom activity (Constructing the ‘square root spiral’): Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion.

Start with a point O and draw a line segment OP₁ of unit length. Draw a line segment P₁P2₂ perpendicular to OP₁ of unit length (see fig.). Now draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in this manner, you can get the line segment P_n-1P_n by drawing a line segment of unit length perpendicular to OP_n-1. In this manner, you will have created the points P₂, P₃, …………….. p_n, …………… and joined them to create a beautiful spiral depicting \(\sqrt{2} \cdot \sqrt{3}, \sqrt{4}, \dots \dots\)
Answer:
Classroom activity :

i) OA = 1 Unit, AB = 1 Unit, ∠A = 90°,
∴OB² = OA² + AB²
= (1)² + (1)²
= 1 + 1
OB² = 2
∴OB = \(\sqrt{2}\)
Similarly, square root spiral can be continued.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.1.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.1

Question 1.
Is a zero a rational number? Can you write it in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0 ? (p, q,ϵ Z, q ≠ 0)
Answer:
Zero is a rational number.
This can be written in the form of \(\frac{p}{q}\) because \(\frac{o}{q}\) is a rational number.
E.g. \(\frac{0}{2}=0, \quad \frac{0}{5}=0\). etc.
Zero belongs to set of rational number.

Question 2.
Find six rational numbers between 3 and 4.
Answer:
We can write six rational numbers between 3 and 4 as
\(3=\frac{21}{7} \text { and } 4=\frac{28}{7}\)
∴ rational numbers between \(\frac{21}{7}\) and \(\frac{28}{7}\).

Question 3.
Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)
Answer:
Rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) are

∴ Rational numbers between \(\frac{30}{50}\) and \(\frac{40}{50}\)

Question 4.
State whether the following statements are true or false. Give reasons for your answers :
(i) Every natural number is a whole number.
Answer:
True. Because set of natural numbers belongs to a set of whole numbers.
∴ W = {0, 1, 2, 3 …………………….}

(ii) Every integer is a whole number.
Answer:
False. Because zero belongs to a set of integers. But -2, -3, -1 are not whole numbers.

(iii) Every rational number is a whole number.
Answer:
False. Because \(\frac{1}{2}\) is a rational number but not a whole number.

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts.

Karnataka SSLC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts

KSEEB SSLC Class 10 Science Chapter 1 Intext Questions

Text Book Part I Page No. 18

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:

We can identify the contents of each test tube, with the help of red litmus paper. Consider the following observations:

1. If there is little change in the colour of red litmus (such as purple), then it is distilled water.
2. If the colour of red litmus remains the same, then it is acid.
3. If the colour of red litmus becomes blue, then it is base.

Text Book Part I Page No. 22

Question 1.
Why should curd and sour substances not be kept in brass and copper vessels?
Answer:
Curd and sour substances contain acid. If we keep curd and sour substances in brass and copper vessel, Nitrogen gas and other harmful products are formed which spoil the food.

Question 2.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:

Hydrogen gas is usually liberated when an acid reacts with a metal. Let us illustrate it with the following examples:

1. Add some pieces of zinc granules into 5ml of a dilute solution of sulphuric acid (H₂SO₄).
2. Shake it well.
3. Pass the produced gas into a soap solution.
4. Now, soap bubbles are formed in the soap solution and these soap bubbles contain hydrogen.
5. Bring a burning candle near a gas-filled bubble. A candle burns with a pop sound. So, the following reaction takes place:
   H₂SO₄(aq) + Zn(s) → ZnSO₄(aq) + H₂↑

Question 3.
Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:

Reaction of zinc granules with dilute sulphuric acid and testing hydrogen Acids Bases and Salts.

Take about 5 ml of dilute sulphuric acid in a test tube and add a few pieces of zinc granutes to it. Pass the gas being evolved through the soap solution. Bubbles are formed in the soap solution, because in soap solution there is hydrogen gas.
H₂SO₄ + Zn ➝ ZnSO₄ + H₂↑

If we take a burning candle near a gas filled bubble, it burns with a pop sound.

Text Book Part I Page No. 25

Question 1.
Why do HCl, HNO₃, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
Hydrogen ions in HCl and HNO₃ are produced in the presence of water. Hydrogen ions cannot exist alone, but they exist after combining with water molecules. Thus hydrogen ions must always be shown as (aq) or hydronium ion (H3O⁺)
The reaction is
HCl_(aq) ➝ H⁺ + Cl^–
H⁺ + H₂O ➝ H₃O⁺
Alcohol and glucose even though they contain Hydrogen they do not show acidic character like HCl and HNO₃.

Question 2.
Why does an aqueous solution of an acid conduct electricity?
Answer:
Due to the presence of hydrogen (H⁺) or hydronium (H₃O⁺) ions, an aqueous solution of an acid conducts electricity.

Question 3.
Why does dry HCl gas not change the colour of the dry litmus paper?
Answer:
Hydrogen ions change the colour of litmus. In dry HCl, there are no H⁺ ions. Ions are formed in aqueous solution. Both HCl and litmus are dry. Hence there is no change in dry HCL and dry litmus paper.

Question 4.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
It is recommended that the acid should be added to water and not water to the acid while diluting an acid because the process is exothermic and lots of heat is generated which may cause the mixture to splash out or cause burns.

Question 5.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer.
When a solution of an acid is diluted, the concentration of hydronium ions (H₃O⁺) per unit volume decreases which also decreases the strength of the acid. The dilution increases the volume of the solution but the concentration of ions remain the same.

Question 6.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
The concentration of hydroxide ions (OH^–) would increase when excess base is dissolved in a solution of sodium hydroxide.
NaOH \(\overset { { H }_{ 2 }O }{ \rightleftharpoons } \) Na⁺ + OH^–
Being a strong base, hydroxide ions are formed on complete dissolution.

Text Book Part I Page No. 28

Question 1.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
If the pH value is less than 7, it represents an acidic solution.
If the pH value is more than 7, it represents a base.
It the pH – 6 is acidic it has more concentration of ions than pH-8 which is a base.

Question 2.
What effect does the concentration of H+(aq) ions have on the nature of the solution?
Answer:
Acids contain more concentration of H⁺ ions. If the concentration is less solution is base.

Question 3.
Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?
Answer:
Yes, the basic solution also has H⁺ ions. However, their concentration is less as compared to the concentration of OH^– ions that makes the solution basic.

Question 4.
Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
If the soil has more acid it is not suitable for agriculture. To increase fertility former should add burnt lime stone or chalk powder.

Text Book Part I Page No. 33

Question 1.
What is the common name of the compound CaOCl₂?
Answer:
Common name of the CaOCl₂ is Bleaching powder.

Question 2.
Name the substance which on treatment with chlorine yields bleaching powder.
Answer:
When chlorine reacts with Ca(OH)₂, bleaching powder is formed.

Question 3.
Name the sodium compound which is used for softening hard water.
Answer:
Na₂CO₃, 10 H₂O is the compound of sodium to soften hard water.

Question 4.
What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.
Answer:

When sodium hydrocarbonate is heated we get sodium carbonate, water and carbon dioxide gas.

Question 5.
Write an equation to show the reaction between Plaster of Paris and water.
Answer:
An equation to show the reaction between plaster of paris and water is

KSEEB SSLC Class 10 Science Chapter 2 Textbook Exercises

Question 1.
A solution turns red litmus blue, its pH is likely to be
a) 1
b) 4
c) 5
d) 10
Answer:
d) 10.
Bases turn red litmus to blue. pH value of 7 is greater than 7. Hence this solution changes red litmus to blue.

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains
a) NaCl
b) HCl
c) LiCl
d) KCl
Answer:
b) the solution contains HCl.

Question 3.
10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount HCl solution (the same solution as before) required to neutralise it will be
a) 4 mL
b) 8 mL
c) 12 mL
d) 16 mL
Answer:
d) 16 mL HCl solution is required.

Question 4.
Which one of the following types of medicines is used for treating indigestion?
a) Antibiotic
b) Anagesic
c) Antacid
d) Antiseptic
Answer:
c) Antacid is used to treat indigestion.

Question 5.
Write word equations and then balanced equations for the reaction taking place when –
a) dilute sulphric acid reacts with zinc granules.
b) dilute hydrochloric acid reacts with magnesium ribbon.
c) dilute sulphuric acid reacts with aluminium powder.
d) dilute hydrochloric acid reacts with iron fillings.
Answer:
(a) H₂SO₄(aq) + Zn(s) → ZnSO₄(aq) + H₂(g)
Sulphuric acid + Zinc ➝ zinc Sulphate + Hydrogen
(b) 2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g)
hydro Chloric acid + magnesium ➝ Magnisium Chloride + Hydrogen
(c) 3H₂SO₄(aq) + 2Al(s) → Al₂(SO₄)₃(aq) + 3H₂(g)
Sulphuric Hydrigen Sulphate + Aluminium ➝ Aluminium + Hydrogen Chloride
(d) 2HCl(aq) + Fe(s) → FeCl₂(aq) + H₂(g)
Hydrochloric acid  + Iron ➝ Ferric + Hydrogen

Question 6.
Compounds such as alchohols and glucose also contain hydrogen but are not categorised as acids. Describe an Activity to prove it.
Solution:
Take solutions of alcohols, glucose in a beaker, take a cork and fix two nails on the cork up to the end. Connect the nails to two terminals of a 6-volt battery through a bulb and a switch. Switch on the current. You will see that bulb does not glow or the current does not pass through the circuit. This means no ions or H⁺ ions are present in the solution. This shows that alcohols and glucose are not acids.

Question 7.
Why does distilled water not conduct electricity, whereas rain water does?
Answer:
Distilled water is pure form of water. This do not contain ions. No electricity flows in distilled water. But rain water is impure and contains ionic groups.
Ex: Acids. Hence rain water is conductor of electricity.

Question 8.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids do not show acidic behaviour in the absence of water because the dissociation of hydrogen ions from an acid occurs in the presence of water only.

Question 9.
Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9, respectively. Which solution is
a) neutral?
b) strongly alkaline?
c) strongly acidic?
d) weakly acidic?
e) weakly alkaline?
Arrange the pH in increasing order of hydrogen-ion concentration.
Answer:
(a) Neutral → Solution D with pH7.
(b) Strongly alkaline → Solution C with pH11.
(c) Strongly acidic → Solution B with pH1.
(d) Weakly acidic → Solution A with pH4.
(e) Weakly alkaline → Solution E with pH9.
The pH can be arranged in the increasing order of the concentration of hydrogen ions as: 11 < 9 < 7 < 4 < 1 or (c) < (e) < (d) < (a) < (b)

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH₃COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?
Answer:
Fizzing will occur more vigorously in test tube. A containing hydrochloric acid. This is because hydrochloric acid is stronger acid than acetic acid and reaction between magnesium ribbon and HCl is faster than between Mg and acetic acid.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.
Answer:
The pH of milk is 6. As it changes into curd, the pH will reduce as during curd formation lactose (sugar) present in milk gets converted into lactic acid. Therefore, the curd is acidic in nature. The acids present in it decrease the pH.

Question 12.
A milkman adds a very small amount of baking soda to fresh milk.
a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
b) Why does this milk take a long time to set as curd?
Answer:
a) Milkman shift the pH of the fresh milk from 6 to slightly alkaline. In this condition milk is not converted into curd as it is in alkaline state.

b) This milk is alkaline. Acids combining with base and becomes neutral. Hence it takes more time to convert into curd.

Question 13.
Plaster of Paris should be stored in a moisture-proof container. Explain why?
Answer:
Plaster of paris should be stored in a moisture proof container. Because plaster of paris absorbs moisture and forms solids. This is called gypsum.

Question 14.
What is a neutralisation reaction? Give two examples.
Answer:
When an acid and base react with each other to give a salt and water, it is termed as neutralisation reaction i.e.,
Acid + Base → Salt + Water
For example:
NaOH + HCl → NaCl + H₂O.
HNO₃ + KOH → KNO₃ + H₂O.

Question 15.
Give two important uses of washing soda and baking soda.
Answer:

Important uses of washing soda are:

1. It is used in glass, soap, and paper manufacturing industries.
2. It is used to remove the permanent hardness of the water.

Important uses of baking soda are:

1. It is used in making baking powder. Baking powder is a mixture of baking soda and a mild acid known as tartaric acid. When it is heated or mixed in water, it releases CO₂ that makes bread or cake fluffy.
2. Therefore, used in baking.
3. It is used in soda-acid fire extinguishers.

KSEEB SSLC Class 10 Science Chapter 2 Additional Questions and Answers

Question 1.
What are olfactory indicators? Give example.
Answer:
Olfactory indicators are the substances whose odour changes in acidic or basic media.
Eg: Vanilla, onion.

Question 2.
Name the different forms of calcium carbonate.
Answer:
Limestone, chalk and marble.

Question 3.
What are alkalies?
Answer:
Bases which are soluble in water are called alkalies.

Question 4.
Name 3 concentrated acids.
Answer:
HCl, H₂SO₄ and HNO₃.

Question 5.
What is the pH range of our body?
Answer:
7.0 to 7.8.

Question 6.
What is acid rain?
Answer:
When pH of rain water is less than 5.6, it is called acid rain.

Question 7.
What are antacids? Give one example.
Answer:
During indigestion the stomach produces too much acid and this causes pain and irritation. To get rid of this pain, people use bases called antacids.
Eg: Magnesium hydroxide (Milk of Magnesia)

Question 8.
Name two sodium salts.
Answer:
Nacl and Na₂SO₄.

Question 9.
What is the formula for bleaching powder?
Answer:
CaOCl₂.

Question 10.
Mention one use of plaster of paris.
Answer:
It is used for making toys.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 7 Control and Coordination are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination.

Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination

KSEEB SSLC Class 10 Science Chapter 7 Intext Questions

Text Book Part I Page No. 85

Question 1.
What is the difference between a reflex action and walking?
Answer:
A reflex action is an automatic reaction for each stimulation in our body initiated by our sense responses e.g., we move our hand immediately after a contact with hot object. It is a direct controlled action. Walking is completely controlled by our brain. On the other hand, is a voluntary action. It requires complete coordination of muscles, bones, eyes etc.

Reflex action	Walking
a) Spinalcordcontrols reflex action.	a) Brain controls walking.
b) It is a spontaneous immediate response to a stimulus. It happens without the will of individual.	b) It is a voluntary action which occurs with the will of individual.

Question 2.
What happens at the synapse between two neurons?
Answer:
Synapse allows delivery of impulses (in chemical form) from neurons to other neurons and the target point, such as muscle cells. Tiny gap between the last portion of axon of one neuron and the dendron of the other neuron is known as a synapse. At a time, it acts as a one way path and transmits impulses in one direction only.

Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
A part of hindbrain-cerebellum is responsible for maintaining posture and equilibrium of the body.

Question 4.
How do we detect the smell of an agarbatti (incense stick)?
Answer:
Forebrain is responsible for thinking work. It has separate areas that are specialized for hearing, smelling, sight, taste, touch etc. The _ forebrain also has regions that collect information or impulses from various receptors. When the smell of an incense stick reaches us, out forebrain detects it. Then, the forebrain interprets it by putting it together with the information received from other receptors and also with the information already stored in the brain.

Question 5.
What is the role of the brain in reflex action?
Answer:
Reflex actions are sudden responses, which do not involve any thinking. A connection of detecting the signal from the nerves (input) and responding to it quickly (output) is called\a reflex arc. The reflex arcs can be considered as connections present between the input and output nerves which meets in a bundle in the spinal cord. The brain is only responsible of the signal and the response.

Text Book Part I Page No. 88

Question 1.
What are plant hormones?
Answer:
Plant hormones or phytohormones are chemicals which are naturally present inside the plants similar to those substances as hormones in animals. These hormones have specific functions and reach the target point to conduct the messages and initiate proper functions like growth, coordinations etc. These are synthesized in one part of the plant body (in minute quantities) and are translocated to other parts when required. The five important phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene.

Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
The movement of shoot towards light is natural and called phototropism. This type of movement is directional and is growth oriented. But, the movement of leaves of the sensitive plants, like Mimosa pudica or “touch me not”, occurs in response to touch or contact stimuli. This movement is not growth oriented.

Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Auxin is a growth-promoting plant hormone.

Question 4.
How do auxins promote the growth of a tendril around a support?
Answer:
Auxin helps the cell grow longer and is synthesized at the shoot tip. When a tendril comes in contact with a support, auxin stimulates faster growth of the cells on the opposite side, so that the tendril forms a coil around the support.

Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
Take two troughs A and B. Fill them with soil. In the trough B place, a small clay pot Now .plant a small seeding in both troughs. Uniformly water the soil in trough A. but pour water in the clay pot of trough B. After a few days take out the seedling from the troughs. The seedling in trough A which used get uniform water has normal straight roots while roots of seedling planted in trough B show bent growth towards the clay pot containing water.

Text Book Part I Page No. 91

Question 1.
How does chemical coordination take place in animals?
Answer:
Hormones cause chemical coordination in animals and help in various functions. Hormone is the chemical messenger that regulates the physiological processes in living organisms. It is secreted by glands. The regulation of physiological processes and control and coordination by hormones comes under the endocrine system. The nervous system along with the endocrine system in our body controls and coordinates the physiological processes.

Question 2.
Why is the use of Iodised salt advisable?
Answer:
If there is deficiency of Iodine we get disease called goitre. Hence use of iodized salt is advisable Iodine is necessary for the thyroid gland to make thyroxin hormone.

Question 3.
How does our body respond when adrenaline is secreted into the blood?
Answer:
Adrenaline is a hormone secreted by the adrenal glands and it is responsible to control any kind of danger or emergency or any kinds of stress. It is secreted directly into the blood and is transported to different parts of the body. It fasten the heartbeat and the breathing rate, and provides more oxygen to the muscles. It also increases the blood pressure. All these responses enable the body to deal with any stress or emergency.

Question 4.
Why are some patients of diabetes treated by giving injections of insulin?
Answer:
Insulin is the hormone secreted by pancreas and helps regulating blood sugar levels. If it is not secreted in proper amounts, the sugar level in the blood rises causing many harmful effects. Hence some patients of diabetes treated by giving injections of Insulin.

KSEEB SSLC Class 10 Science Chapter 7 Textbook Exercises

Question 1.
Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin.
Answer:
(d) Cytokinin.

Question 2.
The gap between two neurons is called a
(a) dendrite
(b) synapse
(c) axon
(d) impulse
Answer:
(b) synapse.

Question 3.
The brain is responsible for
(a) thinking.
(b) regulating the heart beat.
(c) balancing the body.
(d) all of the above.
Answer:
(d) all of the above.

Question 4.
What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
The receptors are usually located in our sense organs such as the inner ear, the nose, the tongue and so on. So gustatory receptors will detect taste while olfactory receptors will detect smell. If these receptors are not working properly, it is harmful to our body.
Eg: If we are suffering from cold, we do not feel the taste of food.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:

Function of a neuron:
At the end of the axon, the electrical impulse sets off the release of some chemicals. These chemicals cross the gap or synapse, and start a similar electrical impulse in a dendrite of the next neuron. A similar synapse finally allows delivery of such impulses from neurons to other cells, such as muscle cells or gland.

Question 6.
How does phototropism occur in plants?
Answer:
The growth movement in plants in response to light stimulus is known as phototropism. The shoots show positive phototropism and the roots show negative phototropism. This means that the shoots bend towards the source of light whereas the roots bend away from the light source. For example: The flower head of sunflower is positively phototropic and hence, it moves from east to west along with the sun.

Question 7.
Which signals will get disrupted in case of a spinal cord injury?
Answer:
The reflex arc connections between the input and output nerves meet in a bundle in the spinal cord. In fact, nerves from all over the body meet in a .bundle in the spinal cord on their way to the brain. In case of any injury to the spinal cord, the signals coming from the nerves as well as the signals coming to the receptors will be disrupted.

Question 8.
How does chemical coordination occur in plants?
Answer:
In animals, control and coordination occur with the help of nervous system. However, plants do not have a nervous system. Plants respond to stimuli by showing movements. The growth, development and responses, to the environment in plants is controlled and coordinated by a special class of chemical substances known as hormones. These hormones are produced in one part of the plant body and are translocated to other needy parts. For example, a hormone produced in roots is translocated to other parts when required. The five major types of phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene. These phytohormones are either growth promoters (such as auxins, gibberellins cytokinins and ethylene) or growth inhibitors such as abscisic acid.

Question 9.
What is the need for a system of control and coordination in an organism?
Answer:
Controlled movement must be connected to the recognition of various events in the enviroment, followed by only the correct movement in response. In other words, living organisms must use systems providing control and coordination. In keeping with the general principles of body organisation. In multi cellular organisms, specilised tissues are used to provide these control and coordination activities. In animals such control and coordination are provided by nervous and muscular tissues.

Question 10.
How are involuntary actions and reflex actions different from each other?
Answer:

Involuntary actions	Reflections
i) These are actions which are not controlled by our will.	There are sudden actions.
ii) These are controlled by medulla of our brain.	These are under the control of Mid brain and hind brain.

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:

Nervous mechanism.	Hormonal mechanism.
1. Information is conducted in the form of electrical impulse	Information is transferred in the form of chemicals called hormones
2. Neurons provide point to point contact for transmission of message	Hormones are carried throughout the body through blood.
3. Message travels rapidly.	Message travels slowly
4. Effect of message continues for a very short period.	Effect of message remains for a longer duration

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
Answer:

Movement in sensitive plants

1. The movement that takes place in a sensitive plant such as Mimosa pudica occurs in response to touch (stimulus).
2. For this movement, the information is transmitted from cell to cell by electric chemical signals as plants do not have any specialised tissue for conduction of impulses.
3. For this movement to occur, the plant cells change shape by changing the amount of water in them.

Movement in our legs

1. Movement in our legs is an example of voluntary actions.
2. The signal or messages for these no action are passed to the brain and hence are consciously controlled.
3. In animal muscle cells, some proteins are found which allow the movement to occur.

KSEEB SSLC Class 10 Science Chapter 7 Additional Questions and Answers

Question 1.
Draw a neat diagram showing Reflex arc and label the parts.
Answer:

Question 2.
Which organ protects the spinal cord and brain?
Answer:
Vertebral column or backbone protects the spinal cord and bony box protects the brain.

Question 3.
Give one example of chemotropism.
Answer:
Growth of pollen tubes towards omles.

Question 4.
Which hormone promote cell division?
Answer:
Cytokinins.

Question 5.
Which hormone helps in regulating blood sugar levels.
Answer:
Insulin.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 7 Control and Coordination will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World.

Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

KSEEB SSLC Class 10 Science Chapter 11 Intext Questions

Text Book Part II Page No. 100

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing focal length or converging power of its lens is called accommodation. The normal eye has a power of accommodation which enables the object as close as 25cm & as far as infinity to be focused on its retina.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
A person with a myopic eye cannot see objects beyond 1.2 m distinctly because, In a myopic eye, the image of a distant object is formed in front of the retina. This defect can be corrected by using a concave lens.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Far point of the human eye with normal vision is near than infinity and near point is 1.2 m.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
A student has difficulty reading the black board while sitting in the last row means he is suffering from myopia. This defect can be corrected using concave lens of suitable power.

KSEEB SSLC Class 10 Science Chapter 11 Textbook Exercises

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) accomodation
(c) near-sightedness
(d) far-sightedness
Answer:
(b) accomodation

Question 2.
The human eye forms the image of an object at its
(a) cornea
(b) Iris
(c) pupil
(d) retina
Answer:
(d) retina

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer:
(c) 25 cm

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina
(c) ciliary muscles
(d) iris
Answer:
(c) ciliary muscles

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his distant vision. For correcting his near vision he needs a lens of power + 1.5 dioptre. What is the focal length of the lens required for correcting (i) distinct vision, and (ii) near vision?
Answer:
i) Lens required for correcting
distant vision = – 5.5
Focal length of lens F \(=\frac{1}{P}\)
\(\mathrm{F}=\frac{1}{-5.5}=0.181 \mathrm{m}\)
Lens required for correcting this defect =-0.181 M
ii) Lens required for correcting near vision = + 1.5 D
Focal length of lens F \(=\frac{1}{P}\)
\(F=\frac{1}{1.5}=0.667 \mathrm{m}\)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
A person is suffering from near vision. So image is formed in front of retina.
Distance of Image, V = – 80 cm
focal length = f
As per lens formula.

f = 80 cm = – 0.8 m,

∴ Power of -125 D is required to correct this problem.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:

Object distance, u = 25 cm
Image distance, v = 1m = -100 m
As per lens formula,

This defect should be corrected by +3.0 power.

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is reached when the object is at a distance of 25 cm from the eye. The focal length of the eye lens cannot be decreased below this minimum limit. Thus an object placed closer than 25cm [or very close to eye] cannot be seen clearly by a normal eye.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
The distance eye lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance inside the eye.

Question 10.
Why do stars twinkle?
Answer:
The twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. This apparent position of the star is not stationary, but keeps on changing slightly, as the physical conditions of the earths atmosphere are not stationary since the stars are very distant, they approximate point – sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the’ apparent position of the star fluctuates and the amount. Of starlight entering the eye flickers some other time, fainter, which is the twinkling effect.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the earth and are thus seen as extended sources. If we consider a plant as a collection of a large number of point sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
Light from the Sun near the horizon passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. However, light from the Sun overhead would travel relatively shorter distance. At noon, the Sun appears white as only a little of the blue and violet colours are scattered. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelength more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark. The sky appears dark to astronaut flying at very high attitude, as scattering is not prominent at such heights.

KSEEB SSLC Class 10 Science Chapter 11 Additional Questions and Answers

Question 1.
Draw a neat diagram showing
(a) neat point of a Hypermetropic eye
(b) Hypermetropic eye
(c) correction for Hypermetropic eye.
Answer:

Question 2.
Which causes presbyopia? Name the lens to correct this defect.
Answer:
This causes due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens.
A common type of bi-focal lense is necessary to correct this defect.

Question 3.
Who discovered spectrum of sunlight?
Answer:
Isaac Newton.

Question 4.
What is the reason for Advance sunrise and delayed sunset?
Answer:
The sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction.

Question 5.
At noon, the sun appears white. Give reason.
Answer:
Because at noon, the sun appears white as only a little of the blue and violet colours are scattered.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 6 Life Processes are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes.

Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes

KSEEB SSLC Class 10 Science Chapter 6 Intext Questions

Text Book Part I Page No. 61

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?
Answer:
Diffusion is a process in which unicellular or small multicellular and generally, aquatic animal’s skin absorb atmospheric oxygen due to concentration difference between internal and external medium of the body. Multicellular organisms like humans possess complex body structure. They have specialised cells and tissues for performing different functions of the body. Special permeable skin is also required for diffusion process. Therefore, diffusion cannot meet huge oxygen requirements of multicellular organisms.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
An indication of life, can be considered as Movement of various kind, breathing or growth etc. However, a living organism can also have movements, which are not visible by naked eye as movements at cellular level. Therefore, the presence of life processes is a fundamental criterion that can be used to decide whether something is alive or not.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Food, water and oxygen are outside raw materials mostly used by an organism. Depending on the complexity of the organism its requirement varies from organism to organism.

An organism needs various raw materials from outside which are as follows:

• Food – To supply energy and provide materials for growth and development of body.
• Water – To provide medium in the cells for all metabolic processes necessary for living condition.
• Oxygen – To oxidise food to release energy.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
The main life processes are nutrition, respiration, transportation, excretion, reproduction, movement etc. which are considered essential.

Text Book Part I Page No. 67

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic Nutrition

1. Food is produced by conversion of atmospheric gases and water in special conditions internally within their body.
2. Presence of chlorophyll is necessary.
3. Food is generally prepared during day time.
4. Almost all plants and some bacteria have this type of nutrition.

Heterotrophic Nutrition

1. Food is taken directly from autotrophs. Then, this food is broken down with the help of enzymes.
2. No pigment is required in this type of nutrition.
3. Food can be prepared at all times.
4. All animals and fungi have this type of nutrition.

Question 2.

Where do plants get each of the raw materials required for photosynthesis?
Answer:

1. Carbon dioxide: Plants get carbon dioxide through stomata.
2. Water: Roots absorb water from soil and transports to leaves.
3. Solar energy: Chlorophyll absorbs solar energy into chemical energy.

Question 3.
What is the role of the acid in our stomach?
Answer:
The hydrochloric acid present in our stomach dissolves bits of food and creates an acidic medium. In this acidic medium, enzyme pepsinogen is converted for pepsin, which is a protein-digesting enzyme.

Question 4.
What is the function of digestive enzymes?
Answer:
The digestive enzymes converts proteins to amino acids, complex carbohydrates into glucose and fat into fatty acids and glycerol.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
The small intestine has millions of tiny finger-like projections called villi. These villi increase the surface area for more efficient food absorption. Within these villi, many blood vessels are present that absorb the digested food and carry it to the blood stream. From the blood stream, the absorbed food is delivered to each and every cell of the body.

Text Book Part I Page No. 71

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
Terristrial animals can breathe the oxygen in the atmosphere, but animals that live in water need to use the oxygen dissolved in water. This oxygen is absorbed by different organs in different animals. All these organs have a structure that increase the surface area which is in contact with the oxygen-rich atmosphere.

Question 2.
What are the different ways in which glucose is oxidised to provide energy in various organisms?
Answer:
In all living organisms, firstly. Glucose is partially oxidised to form two molecules of pyruvate. This process occurs in the cytoplasm of the cell. Further breakdown of pyruvate takes place in different manners in different organisms.

a) Anaerobic respiration: Here the pyruvate is converted into ethanol and carbondioxide in the absence of oxygen. This process takes place in yeast during fermentation. Sometimes, during vigorous muscular activities, when oxygen is inadequate for cellular respiration. Pyruvate is converted into Lactic acid.

b) Aerobic respiration: It takes place in the mitochondria, where puruvate is completely oxidised in the presence of oxygen. At the end CO₂, water and large amount of energy is released.

Question 3.
How is oxygen and carbon dioxide transported in human beings?
Answer:
Haemoglobin transports oxygen molecule to all the cells of body for cellular respiration. The haemoglobin pigment present in the blood gets attached to four O₂ molecules that are obtained from breathing. It thus, forms oxyhaemoglobin and the blood turns oxygenated. This oxygenated blood is distributed to all the body cells by the heart. After giving away O₂ to the body cells, blood takes away CO₂ which is the end product of cellular respiration. Now, the blood becomes de-oxygenated. Since, haemoglobin pigment has less affinity for CO₂, CO₂ is mainly transported in the dissolved form. This de-oxygenated blood gives CO₂ to lung’s alveoli and takes O₂ in return.

Question 4.
How are the lungs designed in human beings to maximise the area for exchange of gases?
Answer:
The exchange of gases takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli. Thus, alveoli are the site for exchange of gases. The lungs get filled up with air during the process of inhalation as ribs are lifted up and diaphragm is flattened. The air that is rushed inside the lungs fills the numerous alveoli present in the lungs. Each lung contains 300-350 millions alveoli. These numerous alveoli increase the surface area for gaseous exchange making the process of respiration more efficient.

Text Book Part I Page No. 76

Question 1.
What are the components of the transport system in human beings? What are the functions of these components?
Answer:
Arteries, veins and capillaries are the components of the transport system in human beings.

1. Arteries: Arteries are the vessels which carry blood away from the heart to various organs of the body.
2. Veins: Veins collect the blood from different organs and bring it back to the heart under high pressure.
3. Capillaries: These are smallest vessels which have one cell thick. Exchange of material between the blood and surrounding cells takes place across this thin wall.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
Separation of pure and impure blood is necessary to keep oxygenated and deoxygenated blood away from mixing. Warm-blooded animals such as birds and mammals maintain a constant body temperature by cooling themselves when they are in a hotter environment and by warming their bodies when, they are in a cooler environment. Hence, these animals require more oxygen (O₂) for more cellular respiration so that they can produce more energy to maintain their body temperature. Such separation helps in a highly efficient supply of oxygen to the body.

Question 3.
What are the components of the transport system in highly organised plants?
Answer:
The transport system in highly organised plants includes xylem and phloem (vascular tissues). Xylem transports water and mineral ions. Phloem conducts food from leaves to other parts of plants.

Question 4.
How are water and minerals transported in plants?
Answer:
The components of xylem tissue (tracheids and vessels) of roots, stems and leaves are interconnected to form a continuous system of water – conducting channels that reaches all parts of the plant. Transpiration creates a suction pressure, as a result of which water is forced into the xylem cells of the roots. Then, there is a steady movement of water from the root xylem to all the plant parts through the interconnected water-conducting channels. Components of xylem tissue helps in the water transport.

Question 5.
How is food transported in plants?
Answer:
Phloem transports food materials to the plant body. The transportation of food in phloem is achieved by utilizing energy stored as ATP. As a result of this, the osmotic pressure in the tissue increases causing water to move into it. This pressure moves the material in the phloem to the tissues which have less pressure. This is helpful in moving materials according to the needs of the plant. For example, the food material, such as sucrose, is transported into the phloem tissue using ATP energy.

Text Book Part I Page No. 78

Question 1.
Describe the structure and functioning of nephrons.
Answer:
Each kidney has large numbers of filtration units called nephrons packed close together.

Functions of nephrons:
Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water, are selectively re-absorbed as the urine flows along the tube. The amount of water re-absorbed depends on how much excess water there is in the body, and on how much of dissolved waste there is to be excreted.

The urine forming in each kidney eventually enters a long tube, the ureter, which connects the kidneys with the urinary bladder. Urine is stored in the urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra. The bladder is muscular, so it is under nervous control. As a result, we can usually control the urge to urinate.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Plants can get rid of excess of water by transpiration. Waste materials may be stored in the cell vacuoles or as gum and resin, especially in old xylem. It is also stored in the leaves that later fall off.

Question 3.
How is the amount of urine produced regulated?
Answer:
Different organisms use varied strategies for excretion. Many unicellular organisms remove the wastes by simple diffusion from the body surface into the surrounding water. Multi cellular organisms use specialised organs to perform the same function. In human beings Nitrogenous waste products are removed by the nephrons in kidneys.

KSEEB SSLC Class 10 Science Chapter 6 Textbook Exercises

Question 1.
The kidneys in human beings are a part of the system for
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) excretion.

Question 2.
The xylem in plants are responsible for
(a) transport of water.
(b) transport of food.
(c) transport of amino acids.
(d) transport of oxygen.
Answer:
(a) transport of water.

Question 3.
The autotrophic mode of nutrition requires
(a) carbon dioxide and water.
(b) chlorophyll.
(c) sunlight.
(d) all of the above.
Answer:
(d) all of the above.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in
(a) cytoplasm.
(b) mitochondria.
(c) chloroplast.
(d) nucleus.
Answer:
(b) mitochondria.

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
Fats are digested in the small intestine. The small intestine gets bile juice and pancreatic juice respectively from the liver and the pancreas. The bile salts (from the liver) break down the large fat globules into smaller globules, so that the pancreatic enzymes can easily act on them. This is referred to as emulsification of fats. It takes place in the small intestine.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
The saliva contains an enzyme called salivary amylase that breaksdown starch which is complex molecule to give simple sugar.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by products?
Answer:
Autotrophs absorbs solar energy and take CO₂ and water, prepare their own food.
Glucose, CO₂ and water are the by products of photosynthesis.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:

The differences between Aerobic and Anaerobic respiration are:

1. Aerobic respiration occurs in the presence of O₂ while anaerobic respiration occurs in the absence of O₂.
2. Aerobic respiration involves the exchange of gases between the organism and the outside environment while in anaerobic respiration exchange of gases is absent.
3. Aerobic respiration occurs in cytoplasm and mitochondria while anaerobic respiration occurs only in cytoplasm.
4. Aerobic respiration always releases CO₂ and H₂O while in anaerobic respiration end products vary.
5. Aerobic respiration yields 36 ATPs while anaerobic respiration yields only 2 ATPs.

Question 9.
How are the alveoli designed to maximise the exchange of gases?
Answer:
The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels.

Question 10.
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Red pigment present in our blood is haemoglobin. It supplies oxygen to all cells of our body Bloodlessness is caused by the deficiency of haemoglobin.

Question 11.
Describe double circulation of blood in human beings. Why is it necessary?
Answer:

A circulation system in which blood pumps through the heart twice during each trip around the body is called double circulatory system. Firstly, blood is pumped into the lungs, where it receives oxygen and becomes oxygenated, and is then pumped back into the heart, before it is finally pumped into the rest of the body.

The human heart has four chambers – the right atrium, the right ventricle, the left atrium and the left ventricle.

Pathway of blood in the heart:

• The heart has superior and inferior vena cava, which carries de- oxygenated blood from the upper and lower regions of the body respectively and supplies this de-oxygenated blood to the right atrium of the heart.
• The right atrium then contracts and passes the de-oxygenated blood to the right ventricle, through an auriculo – ventricular aperture.
• Then, the right ventricle contracts and passes the de-oxygenated blood into the two pulmonary arteries, which pumps it to the lungs where the blood becomes oxygenated. From the lungs, the pulmonary veins transport the oxygenated blood to the left atrium of the heart.
• Then, the left atrium contracts and through the auricular-ventricular aperture, the oxygenated blood enters the left ventricle.
• The blood passes to aorta from the left ventricle. The aorta gives rise to many arteries that distribute the oxygenated blood to all the regions of the body.
• Therefore, the blood goes twice through the heart. This is known as double circulation.

Importance of double circulation:

The separation of oxygenated and de-oxygenated blood allows a more efficient supply of oxygen to every single cells. This efficient system of oxygen supply is very useful in warm-blooded animals such as human beings. As we know, warm-blooded animals have to maintain a constant body temperature. Thus, the circulatory system of humans becomes more efficient because of the double circulation.

Question 12.
What are the differences between the transport of materials in xylem and phloem?
Answer:

Xylem	phloem
i) This tissue trans­ports water and mineral salts.	This tissue trans­ports only food.
ii) Roots absorb water and this is carried to all parts of the plant (upward)	Food prepared in to upwards and downwards in both direction.

Question 13.
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:

1. Structure of alveoli:
   • These are balloon like structures.
   • These provide a surface where the exchange of gases can take place.
     Function:
     Alveoli absorbs oxygen and give up carbon dioxide.
2. Structure of Nephrons:
   Each kidney has large numbers of filtration units called nephrons packed close together.
   Function:
   Blood is reaching kidneys by renal artery. Here filtration takes place. By this glucose, Amino acid and salts are reabsorbed.

KSEEB SSLC Class 10 Science Chapter 6 Additional Questions and Answers

Question 1.
Draw a neat diagram showing schematic representation of transport and exchange of oxygen and carbon dioxide.
Answer:

Question 2.
Write an equation which represents photosynthesis.
Answer:

Question 3.
Mention the function of Lymph.
Answer:
Lymph carries digested and absorbed fat from intestine and drains excess fluid from extra cellular space back into the blood.

Question 4.
Which are Heterotrophic organisms?
Answer:
Animals and Fungi.

Question 5.
Write three steps of photosynthesis.
Answer:
The following events occur during photosynthesis.

1. Absorption of light energy by chlorophyll.
2. Conversion of light energy to chemical energy and splitting of water molecule into hydrogen and oxygen.
3. Reduction of carbon dioxide to carbohydrates.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 6 Life Processes will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment.

Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment

KSEEB SSLC Class 10 Science Chapter 15 Intext Questions

Text Book Part I Page No. 140

Question 1.
What are the trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
Each step or level of a food chain is called Trophic levels.
Example for Food chain

Here grass is a producer because it prepares its own food. This grass is eaten by herbivores means secondary, small carnivores (Frog) are tertiary and higher carnivores are in the fourth level.

Question 2.
What is the role of decomposers in the ecosystem?
Answer:
Microorganisms, comprising bacteria and Fungi, break-down the dead remains and waste products of organisms. These microorganisms are the decomposers as they break-down the complex organic substances into simple inorganic substances that go into the soil and are used up once more by the plants.

Text Book Part I Page No. 142

Question 1.
Why are some substances biodegradable and some non-biodegradable?
Answer:
Substances that are broken down by biological processes are said to be biodegradable. Eg: paper and peel of a fruit. But plastic leather etc. are not broken down. These are called Non-biodegradable.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.
Answer:

1. Leaves of the plants decay and reduces soil fertility,
2. Bio-degradable substances have carbon. When this is burnt, CO₂ and CO are produced and causes air pollution.

Question 3.
Give any two ways in which non-biodegradable substances would affect the environment.
Answer:

1. As these are not decomposing, they cause air pollution and water pollution.
2. Plastic enters stomach of many animals and causes death of animals.

Text Book Part I Page No. 144

Question 1.
What is ozone and how does it affect any ecosystem?
Answer:
Ozone at the higher levels of the atmosphere is a product of UV radiations acting on O₂ molecule. The higher energy UV radiations split apart some molecular Oa in free oxygen (O) atoms. These atoms then combine with the molecular O₂ to form Ozone.

Ozone shields the surface of the earth from ultraviolet (UV) radiation from the Sun. This radiation is highly damaging to organisms for example, it is known to cause skin cancer in human beings.

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
Answer:

1. We must minimise the usage of plastics,
2. We can collect wastes and by this we can produce gas which is an alternate source of energy.

KSEEB SSLC Class 10 Science Chapter 15 Textbook Exercises

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather.
(b) Grass, wood and plastic.
(c) Fruit-peels, cake and lime-juice.
(d) Cake, wood and grass.
Answer:
(b) Grass, wood and plastic.

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango.
(b) Grass, goat and human.
(c) Goat, cow and elephant.
(d) Grass, fish and goat.
Answer:
(b) Grass, goat and human.

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping.
(b) Switching off unnecessary lights and fans.
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above.
Answer:
(d) All of the above.

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
If we kill all the organisms in one trophic level, the population size of organisms in lower level increases uncontrollably and the number of organisms in higher trophic level decreases due to non¬availability of food. This results in an imbalance in ecosystem.

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

• Removing producers: All the heterotrophs die.
• Removing herbivores: Carnivores would not get food.
• Removing carnivores: Herbivores would increase to unsustainable levels.
• Removing decomposers: Organic wastes, plant, and animal dead remains would pile up.
• The role of each and every species belonging to every trophic level is unique.
• No, the organisms of any trophic level cannot be removed without damaging the ecosystem.

Question 6.
What is the biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:
Some harmful chemicals enter our bodies through the food chain, one of the reasons is the use of several pesticides and other chemicals to protect our crops from disease and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals and from the water bodies these are taken up by aquatic plants and animals.

This is one of the ways in which they enter the food chain. This phenomenon is known as biological magnification. This level of magnification be different at different levels of the ecosystem.

Eg: Spraying of DDT will remain for a long time in the environment.

Question 7.
What are the problems caused by the non-biodegradable wastes that we generate?
Answer:

• Non-aesthetic look.
• Death of cattle by ingestion of plastic bags.
• The quality of soil is adversely affected.
• Biomagnification of harmful chemicals like DDT in birds disturb their calcium metabolism.
• Non – biodegradable wastes cause pollution of soil and water.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Answer:
If all the waste we generate is biodegradable, there is a imbalance in nature. Because with the increase of wastes there is decrease in the number of decomposers. These wastes spread every where and microbes are more which causes many diseases to us.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
Damage to the ozone layer causes so many problems. At the higher levels of the atmosphere, ozone performs an essential function. It shields the surface of the earth from ultraviolet radiation from the sun. If ozone layer is damaged no organism can survive. The following are the steps being taken to limit this damage.

1. We should minimize the use of vehicles.
2. We should not encourage the burning of fossilic fuels.
3. It is now mandatory for all the manufacturing companies to make CFC- free refrigerators throughout the world.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment, drop a comment below and we will get back to you at the earliest.