KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction.

Karnataka SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

KSEEB SSLC Class 10 Science Chapter 10 Intext Questions

Text Book Part II Page No. 78

Question 1.
Define the principal focus of a concave mirror.
Answer:
The number of rays parallel to the principal axis are falling on a concave mirror which meat at a point is called principal focus of the concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 2f Here R = 20 cm
20 = 2f
∴ \(f=\frac { 20 }{ 2 } =10\)
∴ Focal length = 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Because these mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving.

Text Book Part II Page No. 81

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature, R = 32 cm
Radius of curvature = 2f
\(R=2f=\frac { R }{ 2 } =\frac { 32 }{ 2 } =16\)
∴ Convex mirror focal length is = 16cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
\(M=\frac { Height\quad of\quad image }{ Height\quad of\quad object }\)
\(=\frac{h_{1}}{h_{0}}=\frac{-u}{v}\)
Let the height of object be h then height of image h = – 3h
\(=\frac{3 h}{h}=\frac{-v}{u}=\frac{v}{u}=3\)
∴ Distance of object, u = – 10 cm
v = 3 × (10) = – 30 cm
Here – sign indicates, image is real and it is 30 cm in front of concave mirror.

Text Book Part II Page No. 86

Question 1.
A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Lightray bend towards normal. Because when a ray of light enters from rearer medium to denser medium, it changes its direction in the second medium.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vaccum is 3 108 ms-1
Answer:
Refractive index, nm
\(=\frac { Velocity\quad of\quad light\quad in\quad vaccum\quad }{ Refractive\quad Index\quad of\quad glass } \)
\(=\frac{3 \times 10^{8}}{1.50}=2 \times 10^{8} \mathrm{m} / \mathrm{s}\)

Question 3.
Find out, from Tabel 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
Diamond is having highest optical density.
Air is having lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table 10.3
Answer:
Light travel faster in water because Refractive index of water is lesser than kerosene and turpentine.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means Ratio of velocity of light in air and velocity of air in diamond is 2.42.

Text Book Part II Page No. 94

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of lens whose focal length is 1 metre 1 D = 1 m-1

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Image of Needle is real and inverted means this is real image it is 2f
Image is at a distance of 50 cm
Hence needle is kept 50 cm in front of convex lens.
Distance of object, u = – 50 cm.
Distance of image v = 50 cm
Focal length f = ?
As per lens formula.
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction 94 Q 2.1
f = 25 cm = 0.25 m
Power of the lens
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction 94 Q 2
Power of the lens P = + 4D.

KSEEB SSLC Class 10 Science Chapter 10 Textbook Exercises

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) At twice the focal length.

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
(a) both concave.

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
(c) A convex lens of focal length 5cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror?
What is the nature of the image? Is the image larger or smaller than the object?
Draw a ray diagram to show the image formation in this case.
Answer:
Distance of the object = o to 15 cm
Nature of image = virtual, erect and bigger than object
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 7

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.
(b) A convex mirror is used as a side/ rear-view mirror of a vehicle because,

  • A convex mirror always forms an erect, virtual, and diminished image of an object placed anywhere in front of it.
  • A convex mirror has a wider field of view than a plane mirror of the same size.

(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
This lens gives full image, though one-half of this lens is covered with black paper as shown in below figure.
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 9
As shown in figure light ray moves in half part and image is formed in another part of the lens.
If black paper is covered in lower part: Following figure explain this
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 9

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Height of object, h = 5 cm
Distance of object from converging lens u = 25 cm
Focal length of lens f = 10 cm
As per lens formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10
Converging lens, \(\frac{h_{1}}{h_{0}}=\frac{v}{u}\)
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10.3
= – 3.3 cm
Images is inverted and it is formed it is formed behind the lens about 16.7 cm. Its height is 3.3 cm.
Diagram is as follows:
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10.2

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Focal length (F1) of concave lens
f = 15 cm
Image distance, v = – 10 cm
As per lens formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 11
u = -30 cm
Negative sign indicates, image is front of the lens about 30 cm.
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 11.1

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror,
f = +15 cm
Object distance, u = -10 cm
As per lens formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 12
Magnification \(=\frac{v}{u}=\frac{-6}{-10}=0.6\)
Virtual image is formed at the distance of 6 cm and it is erect.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The positive [+] sign of magnification [m] indicates that the image is virtual and erect. The magnification m = 1 indicates that the image is of the same size as the object. Thus, the magnification of +1 produced by a plane mirror means the image formed in a plane mirror is virtual, erect and of the same size as the object.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = 20 cm
Height of object h = 5 cm
Radius of curvature R = 30 cm
R = 2f, f = 15 cm
As per mirror formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 14
Positive sign indicates image is formed behind the mirror
Magnification \(=\quad \frac { Image\quad distance }{ Object\quad distance } \)
\(=\frac{-8.57}{-20}=0.428\)
Image is behind the mirror because magnification is positive
Magnification \(=\frac{\text { Image distance }}{\text { Object distance }}\)
\(=\frac{h^{1}}{h}\)
h1 = m × h = 0.428 × 5 = 2.14 cm

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focussed image can be obtained? Find the size and nature of the image.
Answer:
Objective distance, u = 27 cm
Object height, h = 7 cm
Focal length, f = 30 cm
R = 2f, f = -18 cm
As per mirror formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15
Screen should be placed in front of mirror at the distance of = 54 cm
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15.1
Negative sign of magnification indicates image is real Magnification,
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15.2
h1 = 7 × (2) = -14 cm
Image is inverted because of negative sign.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Power of lens, P \(=\frac{1}{f}\)
P = -2D
f \(=\frac{-1}{2}=-0.5 \mathrm{cm}\)
Negative sign indicates this is concave lens.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Power of lens, P \(=\frac{1}{f}\)
P = 1.5 D
F \(=\frac{1}{1.5}=\frac{10}{15}=0.66 \mathrm{m}\)
This is converging lens means convex lens.

KSEEB SSLC Class 10 Science Chapter 10 Additional Questions and Answers

Question 1.
What are spherical mirrors?
Answer:
Mirrors, whose reflecting surfaces are spherical, are called spherical mirrors.

Question 2.
What is pole of the mirror?
Answer:
The centre of the reflecting surface of a spherical mirror is a point called the pole.

Question 3.
What is principal axis?
Answer:
The line passing through the pole and the centre of curvature of a spherical mirror is called principal axis.

Question 4.
Draw a ray diagram of concave mirror and convex mirror.
Answer:

KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ad Q 5.1
(a) Concave mirror (b) Convex mirror

Question 5.
Draw a neat diagram showing Refraction of light through a rectangular glass slab
Answer :
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction ad Q 5

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KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current.

Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

KSEEB SSLC Class 10 Science Chapter 13 Intext Questions

Text Book Part I Page No. 118

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle get deflected when brought near a bar magnet because a compass needle is in fact, a small bar magnet. The ends of the compass needle point approximately towards north and south directions.

Text Book Part I Page No. 122

Question 2.
Draw magnetic Held lines around a bar magnet.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 122 Q 1
Field lines around a bar magnet

Question 3.
List the properties of magnetic field lines.
Answer:
The properties are:

  • They travel from north pole to south pole outside the magnet and south pole to north pole inside the magnet.
  • They are closed and continuous curves.
  • Two magnetic field lines never intersect each other. If the lines intersect, then at the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic fields which is not possible.
  • The number of field lines per unit area is the measure of the strength of magnetic field, which is maximum at poles. The magnetic field is strong, where the field lines are close together and weak where the lines are far apart.

Question 4.
Why don’t two magnetic field lines intersect each other?
Answer:
Two magnetic fields lines of force never intersect each other. If the lines intersect, then at [the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic field, which is not possible.

Text Book Part I Page No. 123, 124

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 123,124 Q 1
Since the current passes through the loop in a clockwise direction, therefore the front face of the loop will be the south pole and the back face, ie, the face touching the table will be north pole. According to right-hand rule, the direction of the magnetic field inside the loop will be pointing downward. Outside the loop, the direction of the magnetic field will be upward.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 123,124 Q 2
The figure indicates that the magnetic field is the same at all points in the solenoid. That is field is uniform inside the solenoid.

Question 3.
Choose the correct option.
The magnetic field inside a long straight solenoid-carrying current
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:
(d) is the same at all points.

Text Book Part I Page No. 125, 126

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity
(d) momentum.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if

  1. current in rod AB is increased;
  2. a stronger horse-shoe magnet is used; and
  3. length of the rod AB is increased?

Answer:

  1. displacement of A is increased.
  2. If a stronger horse-shoe magnet is used magnetic field is increasing.
  3. current flows is more.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward.
Since the positively charged particle alpha particle projected towards west, so the direction of current is towards west. Now the deflection is towards north, so the force is towards north. Now hold the forefinger, centre finger and thumb of our left – hand at right angles to one another. Let us adjust the hand in such a way that our centre finger points towards west and thumb points towards north. If we look at our forefinger, it will be pointing, upward. Thus, the magnetic field is in the upward direction. So, the correct answer is (d).

Text Book Part I Page No 127

Question 1.
State Fleming’s left-hand rule.
Answer:
According to this rule, stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the first finger points in the direction of the Magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
A current-carrying conductor when placed in a magnetic field experiences a force. This is the principle of an electric motor.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring reverse the direction of current in the armature coil after every half rotation i.e., it acts as a commutator. The reversal of current reverses, the direction of the forces acting on the two arms of the armature after every half rotation. This allows the armature coil to rotate continuously in the same direction.

Text Book Part I Page No. 130

Question 1.
Explain different ways to induce current in a coil.
Answer:
We can induce current in a coil either by moving it in a magnetic field or by changing the magnetic around it. It is convenient in most situations to move the coil in a magnetic field.

Text Book Part I Page No. 131

Question 1.
State the principle of an electric generator.
Answer:
Based on the phenomenon of electromagnetic induction, electric generator are prepared. In an electric generator, Mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. This is the principle of an electric generator.

Question 2.
Name some sources of direct current.
Answer:
Dry cell, Battery and D.C. generator.

Question 3.
Which sources produce alternating current?
Answer:
A.C. generator and D.C. generator.

Question 4.
Choose the correct option. A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution.
Answer:
(c) half revolution.

Text Book Part I Page No. 132

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

  1. Electric fuse
  2. Earthing wire.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
P = VI
Here P = 2 KW = 2000 W
V = 220
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 132 Q 2

The current drawn by this electric oven is 9 A whereas the fuse in the circuit is ( only 5 A capacity. When a high current of 9 A flows through the 5 A fuse, the fuse wire will get heated too much, melt and break, the circuit. Therefore, when a 2 kW power rating electric oven is operated in a circuit having a 5 A fuse will blow off cutting off the power supply in this circuit.

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:

  1. Each appliance has a separate switch to ON/OFF the flow of current through it.
  2. The use of an electric fuse prevents the electric circuit and the appliance from possible damage by stopping the flow of unduly high electric current.
  3. We should not connect too many appliances to a single socket to prevent overloading.

KSEEB SSLC Class 10 Science Chapter 13 Textbook Exercises

Question 1.
Which of the following correctly describes the magnetic Held near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.

Question 2.
The phenomenon of electro-magnetic induction is
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electro-magnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(c) AC generator will generate a higher voltage.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.
Answer:
(a) False
(b) true
(c) true
(d) False.

Question 7.
List two methods of producing magnetic fields.
Answer:

  1. Permanent magnet
  2. Electromagnet.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 8

One end of the solenoid behaves as a magnetic north pole, while the other behaves as the south pole. The field lines inside the solenoid are in the form of parallel straight lines. This indicates that the magnetic field is the same at the points inside the solenoid.

As shown in figure a strong magnetic field produced inside a solenoid can be used to magnetise a piece of Magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
If the direction of magnetic field and flow of electric current are mutually perpendicular then force experienced by a current-carrying conductor in a magnetic field is largest.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
According to Fleming’s left-hand rule, the magnetic field acts in the vertically downward direction.
Note that the direction of current will be opposite to that of the electron beam.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 11

Principle:
An electric motor is a rotating device that converts electrical energy to mechanical energy working.

Working:
Current in the coil ABCD enters from the source battery through conducting brush X and flow back to the battery through brush Y. Notice that the current in the Arm AB of the coil flows from A to B. In arm CD it flows from C to D that is opposite to the direction of current through arm AB on applying Fleming’s left hand rule for the direction of force on a current¬carrying conductor in a magnetic field. We find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards.

Thus the coil and the Axle O, mounted free to turn about an axis, rotate anti-clockwise at half rotation. Q makes contact with the brush X and P with brush Y. Therefore the current in the coil gets reversed and flows along the path DCBA. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down, is now pushed up and the arm CD previously pushed up is pushed down. There is a continuous rotation of the coil and to the axle.
Split rings in electric motors acts as a commutator.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motor is used as an important component in electric fans, refrigerators, mixers, washing machines, computers, MP3 players etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

  1. pushed into the coil
  2. withdrawn from inside the coil
  3. held stationary inside the coil?

Answer:

  1. There is a momentary deflection in the needle of the galvanometer.
  2. Now the galvanometer is deflected towards the left showing that the current is now set up in the direction opposite to the first.
  3. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in coil B? Give reason.
Answer:
If the current in the coil A is changed there is a change in its magnetic field. By this electricity is induced in B. This is called Electromagnetic induction.

Question 15.
State the rule to determine the direction of a

  1. magnetic field produced around a straight conductor carrying current,
  2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
  3. current induced in a coil due to its rotation in a magnetic field.

Answer:
(i) Right-hand thumb rule: If the current-carrying conductor is held in the right hand such that the thumb points in the direction of the current, then the direction of the curl of the fingers will be given the direction of the magnetic field.

(ii) Fleming’slefthandrule: Stretch the forefinger, the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger in the direction of the current, then the thumb points in the direction of a force in the conductor.

(iii) Fleming’s right-hand rule: Stretch the thumb/ forefinger and the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, thumb in the direction conductor, then the central finger points in the direction of current induced in the conductor.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 16

Principle: In an electric generator, mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. It is working on the principle of electromagnetic induction.

Working: When the Axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Let us say the coil ABCD is rotated clockwise in the arrangement. By applying Fleming’s right-hand rule, the induced currents are set up in these arms along with the directions AB and CD. Thus an induced current flows in the direction ABCD. If there are a larger number of turns in the coil, the current generated in each turn adds up to give a large current through the coil. This means that the current in the external circuit flows from B2 and B1.

After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. The current in the external circuit now flows from B1 to B2. Thus after every half rotation the polarity of the current in the respective arms changes.

There are two brushes and in the electric generator, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down. Because of these Brushes unidirectional current is produced.

Question 17.
When does an electric short circuit occur?
Answer:
Overloading can occur when the live wire and the neutral wire come into direct current (This occurs when the insulation of wires is damaged or there is a fault in the appliance) In such a situation, the current in the circuit abruptly increases. This is called short-circuiting.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
This is used as a safety measure, especially for those appliances that have a metallic body, for example, electric press, toaster, table fan, refrigerator, etc. The metallic body is connected to the earth wire which provides a low-resistance conducting path for the current. Thus earth wire ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth and the user may not get a severe electric shock.

KSEEB SSLC Class 10 Science Chapter 13 Additional Questions and Answers

1. Fill in the blanks:

Question 1.
Magnetic field is a quantity that has both …….. and ……
Answer:
Magnitude, direction.

Question 2.
An electric current through a metallic conductor produces a ……. around it.
Answer:
Magnetic field.

Question 3.
In electric motors, the …… acts as a commutator
Answer:
split ring.

Question 4.
…… current reverses its direction periodically.
Answer:
Alternating.

Question 5.
In our country the potential difference between two wires is ……
Answer:
220 V.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2

(Unless stated otherwise, take n \(=\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ‘n’.
Solution:
Radius of hemisphere, r = 1 cm.
Volume of hemisphere, V \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2 \pi}{3} \times(1)^{3}\)
Radius of Cone, r = 1 cm,
Height of Cone, h = 1 cm.
∴ Volume of a cone, V \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 1
∴ Volume of Total cube \(=\frac{2 \pi}{3}+\frac{\pi}{3} \mathrm{cm}^{3}\)
= πcm3

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2
solution:
Volume of cylinder, V = πr2h
= π × (1.5)2 × 8
= 18π cm3.
Volume of Cylinder V \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \pi \times(1.5)^{2} \times 2\)
\(=\frac{3}{2} \pi \mathrm{cm}^{3}\)
∴ The volume of air contained in the model that Rachel made.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2.1
= 66 cm3.

Question 3.
A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see figure given aside).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.1
Solution:
Length of cylindrical Gulab jamun,
l = 5 – (1.4 + 1.4) = 2.2 cm.
Radius, r = 1.4 cm.
∴ Volume of Cylinder (1 Gulab Jamun), = Left hemisphere + Volume of Cylinder + Right hemisphere
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.2
∴ Total volume of such 45 Gulab jamun,
\(=45 \times \frac{22}{3} \times 0.28 \times 12.2\)
= 22 × 0.28 × 183
= 1127.28 cm3.
∴ Gulab Jamun contains sugar syrup upto about 30% of its volume.
∴ \(1127.28 \times \frac{30}{100}\)
= 338 cm3.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see
Figure)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4
Solution:
Radius of conical depression, r = 0.5 cm.
(Depth) height, h = 1.4 cm.
Volume of 1 depression which is conical, \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4.1
Volume of such 4 depressions
\(=4 \times \frac{11}{30} \mathrm{cm}^{3}\)
\(=\frac{44}{30} \mathrm{cm}^{3}\)
∴ Total volume of wooden pen-stand, \(=(15 \times 10 \times 3.5)-\frac{44}{30}\)
= 525 – 1.47
= 523.53 cm3

Question 5
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Volume of water in a Cone,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 5
Volume of lead shot \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \pi \times(0.5)^{3}\)
\(=\frac{\pi}{6} \mathrm{cm}^{3}\)
Let the number of lead balls kept in vessel \(\frac{1}{4}\) of the water flows out,
\(\Rightarrow n \times \frac{\pi}{6}=\frac{1}{4} \times \frac{200 \pi}{3}\)
\(\Rightarrow n \times \frac{\pi}{6}=\frac{100}{6} \pi\)
∴ n = 100
∴ Number of lead balls =100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. (Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14).
Solution:
Height of solid iron pole, h = 220 cm.
Base diameter of pole, d = 24 cm.
∴ Radius, r = 12 cm.
Height of cylinder, h = 60 cm.
Radius of cylinder, r= 8 cm.
Mass of the pole = ?
Volume of 1st Cylinder, V = πr2h
= π × (12)2 × 220
Volume of 2nd cylinder, V = πr2h
= 77 × (8)2 × 60
∴ Total Volume = π × (12)2 × 220 + π × (8)2 × 60
= {144 × 220 + 64 × 60}π
= 35520π
= 111532.8 cm3.
Mass of iron about 1 ccm is 8 gm.
∴ Mass of iron 111532.8 ……. ??
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 6
= 111.5328 × 8 kg.
= 892.26 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm. and its height is 180 cm.
Solution:
(i) Radius of cylinder GHEF, r = 60 cm.
Height, h = 120 + 60
= 180 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7
Volume of cylinderical vessel \(=\pi r^{2} h\)
= π × (60)2 × 180
(ii) Volume of hemisphere + Volume of cone
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.1
(iii) Quality of water remained in cylinder.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.2

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and n = 3.14.
Solution:
Diameter of spherical glass vessel = 8.5 cm.
∴ r = 4.25 cm.
Length of cylindrical neck = 8 cm.
Diameter = 2 cm.
∴ Radius, r = 1 cm.
Volume of water she found = 345 cm3.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8
(i) Volume of Cylinder neck, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.1
(ii) Volume of cylindrical vessel \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \pi \times(4.25)^{3}\)
∴ Total volume of water, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.2
∴ Her answer is not correct.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4

Question 1.
The following distribution gives the dialy income of 50 workers of a factory.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.1
n = 50, ∴ \(\frac{n}{2}\) = 25
On a graph paper mark the following points :
(120, 12), (140, 26), (160, 34), (180, 40), (200, 50).
For the Ogive graph,
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.2

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.1
To draw ogive of the less than type.
we have to join the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
From the graph, Median is 46.5 kg
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.2
n = 35 ∴ \(\frac{\mathrm{n}}{2}\) = 17.5
Class interval which has median is = (46 – 48)
l = 46, n = 35, f = 14, cf = 14, h = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 2.3
= 46.5
∴ Median = 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.1
We can draw Ogive graph by plotting ordered pairs:
(50, 100), (55, 98), (60, 90), (65, 78). (70, 54), (75, 16).
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.2

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.1
(i) n = 68, ∴ \(\frac{n}{2}\) = 34
The median is in class interval (125 – 145)
l = 125, n = 68, f = 20, cf = 22, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.2
= 125 + 12
∴ Median = 137 units.

(ii) To find out Mode:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.3
Here, Mode is in class interval (125 – 145)
Maximum frequency, l = 125, f<sub>1</sub> = 20, f<sub>0</sub> = 13, f<sub>2/sub> = 14, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.4
= 125 + 10.76
= 135.76
∴ Mode = 135.76 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of ‘x’ and ‘y’
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2
Solution:
Median is 28.5
Class interval which has median is = (20 – 30)
l = 20, n = 60, f = 20, cf = 5 + a × h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2.1
17 = 25 – x
∴ x = 8
5 + x + 20 + 15 + y + 5 = 60
x + y – 45 = 60
x + y = 15
y = 15 – x
y = 15 – 8
∴ y = 7
∴ x = 8, y = 7

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.1
n = 100 ∴ \(\frac{n}{2}\) = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.2
= 35 + 0.76
∴ Median = 35.76 years

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, 171.5 – 180.5)
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.1
n = 40 ∴ \(\frac{n}{2}\) = 20
C.I. which has median is (144.5 – 153.5)
l = 144.5, n = 40, f= 12, cf = 17, h = 9
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.2
= 144.5 + 2.25
∴ Median = 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5
Find the median life time of a lamp.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.1
n = 400 ∴ \(\frac{n}{2}=\frac{400}{2}=200\)
Class interval having median is = (3000 – 3500)
l = 3000, n = 400, f = 86, cf = 130, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.2
= 3000 + 406.98
∴ Median = 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.1
(i) n = 100, ∴ \(\frac{n}{2}\) = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.2
= 7 + 1.05
∴ Median = 8.05 Letters

(ii) Clall interval which has mode is (7 – 10)
Maximum frequency, l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.3
= 7 + 0.88
∴ Mode = 7.88

(iii) Mean(\(\overline{X}\)) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.4
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.5
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32
∴ (i) Median = 8.05 letters
(ii) Mode = 7.88
(iii) Mean = 8.32.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.1
(i) n = 30, ∴ \(\frac{\mathbf{n}}{2}\) = 15
Class interval having median is (55 – 60)
l = 55, n = 30, f = 6, cf = 13, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.2
= 55 + 1.67
∴ Median = 56.67 kg.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds.

Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

KSEEB SSLC Class 10 Science Chapter 4 Intext Questions

Text Book Part II Page No. 5

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO2?
Answer:
In carbon dioxide molecule, the two oxygen atoms are bonded on either side with carbon atom be double bonds. These there are 2 double bonds in CO2. Carbon shares its electrons in the formation of a double bond with one
oxygen atom and another two electrons with another oxygen atom. In this process, both the oxygen atoms and the carbon atom acquire the stable electronic configuration of the noble gas neon. The formation of CO2 molecule is shown below.
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 5 Q 1

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?
(Hint – The eight atoms of sulphur are joined together in the form of a ring.)
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 5 Q 2

Text Book Part II Page No. 12

Question 1.
How many structural isomers can you draw for pentane?
Answer:
Pentane has 3 structural isomers. We can write as follows.
i) CH3-CH2-CH2-CH2-CH3
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 12 Q 1

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

  1. Catenation: It is the ability to form bonds with other atoms of carbon.
  2. Tetravalency: With the valency of four, carbon is capable of bonding with four other atoms.

Question 3.
What will be the formula and electron dot structure of cyclopentane?
Answer:
Molecular formula of cyclopentane is: C5H10
Electron dot structure:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 12 Q 3

Question 4.
Draw the structures for the following compounds.

  1. Ethanoic acid
  2. Bromopentane*
  3. Butanone
  4. Hexanal.

* Are structural isomers possible for bromopentane ?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds tableee

Question 5.
How would you name the following compounds?
(i) CH3—CH2—Br
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 12 Q 5 1
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds 12 Q 5 2
Answer:

  1. Bromoethane
  2. Methanol
  3. Hexane.

Text Book Part II Page No. 15

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
Addition reaction means adding oxygen. Adding ethanol to potassium permanganate, we get ethanoic acid. Hence this reaction is called oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
Air, also contains other gases like nitrogen, carbon dioxide and few more gases apart from oxygen. When ethyne is burnt in air, it gives a sooty flame. This is due to incomplete combustion caused by the limited supply of oxygen. However, if ethyne is burnt with oxygen, it gives a clean flame with temperature 3000°C because of complete combustion. This oxy-acetylene flame is used for welding. It is not possible to attain such a high temperature without mixing oxygen. This is the reason why a mixture of ethyne and air is not used,
2HC ≡ CH + 5O2 → 4CO2 + 2H2O + Heat.

Text Book Part II Page No. 18

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
All the carboxylic acids decompose sodium hydrogen carbonate giving brisk effervescence of carbon dioxide gas whereas ethanol does not react with sodium hydrogen carbonate

Experiment:

  1. Take two test tubes, label them as A and B
  2. Take about 0.5 g of sodium hydrogen carbonate (NaHco3) in each test tube
  3. Add 2 ml of ethanol in test tube A and 2ml of ethanoic acid in test tube B.
  4. We can observe the gas bubbles in test tube B. No such bubbles are seen in test tube A. Pass the gas produced in test tube B through lime water taken in another test tube
  5. We will find that lime water turns milky It is a test for carbon dioxide.

Hence, this experiment proves that when ethanoic acid reacts with sodium hydrogen carbonate, then carbon dioxide gas is produced with an effervescence (a rapid evolution of gas bubbles). Ethanol does not react with NaHCO3.

Question 2.
What are oxidising agents?
Answer:
Oxidising agents are the substances that gain electrons in redox reaction and whose oxidation number is reduced. Examples: KMnO4 or K2Cr2O7. They have the ability to oxidize or give their oxygen to other substances.

Text Book Part II Page No. 20

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
Detergent gives lather both with hard and soft water, while a soap gives lather with soft water only. Thus, it is not possible to check if the water is hard by using a detergent.

Question 2.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
A soap molecule has two parts namely hydrophobic and hydrophilic. With the help of these particles, it attaches to the grease or dirt particle and forms a cluster called micelle. These micelles remain suspended as a colloid. To remove these micelles, it is necessary to agitate clothes.

KSEEB SSLC Class 10 Science Chapter 4 Textbook Exercises

Question 1.
Ethane, with the molecular formula C2H6 has
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.
Answer:
(b) 7 covalent bonds.

Question 2.
Butanone is a four-carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone.
(d) alcohol.
Answer:
(c) ketone.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH3Cl.
Answer:
Carbon has a valency of four. It shares one electron to each 3 hydrogen atoms and one more electron with chlorine.
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 4

Question 5.
Draw the electron dot structures for
(a) ethanoic acid.
(b) H2S.
(c) propanone.
(d) F2.
Answer:
a) Ethanoic acid
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 5 1

b) H2S
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 5 2

c) propanone
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 5 3

(d) F2
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 5 4

Question 6.
What is an homologous series? Explain with an example.
Answer:

A homologous series is a series of carbon compounds that have different numbers of carbon atoms but contain the same functional group. Every next member of a homologous series has a clear difference of 14 units of mass.

For example, methane, ethane, propane, etc., are all part of the alkane homologous series. The general formula of this series is CnH2n+2.

An example is explained with formula as below:

  1. Methane, CH4
  2. Ethane, CH3CH3
  3. Propane, CH3CH2CH3
  4. Butane, CH3CH2CH2CH3

It can be noticed that there is a difference of -CH2 unit between each successive compound.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:

Ethanol and Ethanoic acid can be differentiated on the basis of their following properties by:

  1. Ethanol is a liquid at room temperature with a pleasant smell. Ethanoic acid has a melting point of 17°C. Since it is below the room temperature so, it freezes during winter. Moreover, ethanoic acid has a smell like vinegar.
  2. Ethanol does not react with metal carbonates while, ethanoic acid reacts with metal carbonates to form a salt, water and carbon dioxide.
    For example:
    2CH3COOH + Na2CO3 → 2CH3COONa + CO2 +H2O
  3. Ethanol does not react with NaOH while ethanoic acid reacts with NaOH to form sodium ethanoate and water.
    For example,
    CH3COOH + NaOH → CH3COONa + H2O
  4. Ethanol is oxidized to give ethanoic acid in the presence of acidified KMnO4 while no reaction takes place with ethanoic acid in the presence of acidified KMnO4.

Difference in physical properties:

Ethanol Ethanoic acid
This is in liquid form at room temperature. Its melting point is 156° K. Its melting point is 290K and hence it often freezes during winter in cold climates.
Difference in chemical properties
Ethanol will not react with metallic carbo­nates. Ethanoic acid reacts with carbo­nates and Hydrogen carbonate and forms salts, carbon dioxide and water.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
A soap molecule has two ends. One end is hydrophilic and another end is hydrophobic. When soap is dissolved in water and clothes are put in the soapy solution, soap molecules converge in a typical manner to make a structure is called micelle. The hydrophobic ends of different molecules surround a particle of grease and make the micelle, which is a spherical structure.

In this, the hydrophilic end is outside the sphere and hydrophobic end is towards the centre of the sphere. This is why micelle formation takes place when soap is added to water. Since ethanol is not as polar as soap, micelles will not be formed in other solvents such as ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon in all its allotropic forms, burns in oxygen to give carbondioxide along with the release of heat and light. Most carbon compounds also release a large amount of heat and light on burning. Hence carbon and its compounds are used as fuels for most applications.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hard water often contains salts of calcium and magnesium. Soap molecules react with the salts of calcium and magnesium and form a precipitate. This precipitate begins floating as an off-white layer over water. This layer is called scum. Soaps lose their cleansing property in hard water because of the formation of scum.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Soap is basic in nature, hence red litmus changes to blue. Blue litmus is seen blue only.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:

Hydrogenation is a reaction between hydrogen and other compounds in the presence of the desired catalyst. Hydrogenation is used for reducing saturated hydrocarbons. Hydrogenation is an addition reaction. For example: When ethane is heated with the catalyst, nickel, it is reduced to ethane.

Industrial application:

  1. In the petrochemical industry, hydrogenation is used to convert alkenes into alkanes (paraffin) and cyclo-alkanes.
  2. It is also used to prepare vegetable cooking fat from vegetable oils.

Question 13.
Which of the following hydro-carbons undergo addition reactions:
C2H6, C3H8, C3H6, C2H2 and CH4.
Answer:
C2H6 and C2H2 are unsaturated Hydrocarbons. Hence these undergo addition reactions.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Butter contains saturated fats. Therefore, it cannot be hydrogenated. On the other hand, oil has unsaturated fats. That is why it can be hydrogenated to saturated fats (solids).

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Ex Q 15
Soap are molecules in which the two ends have differing properties. One is hydrophilic, that is, it interacts with water, while the other end is hydrophobic, that is, it interacts with hydrocarbons. In the clusters of molecules in which the hydrophobic tails are on the surface of the cluster. This formation is called micelle.

Since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus the dirt suspended in the micelles is also easily rinsed away.

KSEEB SSLC Class 10 Science Chapter 4 Additional Questions and Answers

Question 1.
Write the electron dot formula of O2.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Add Q 1

Question 2.
What is substitution Reaction? Give an example.
Answer:
If one type of atom or a group of atoms takes the place of another, it is called substitution reaction.
Eg: CH4 + Cl2 CH3Cl + HCl (in the presence of sunlight)

Question 3.
Name 2 commercially important compounds.
Answer:
Ethanol and ethanoic acid.

Question 4.
Give an example for Esterification reaction.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Add Q 4

Question 5.
Write one use of ester.
Answer:
Esters are used in making perfumes and as flourishing agents.

Question 6.
What are detergents?
Answer:
Detergents are generally sodium salts of sulphonic acids or ammonium salts with chlorides or bromides etc.

Question 7.
Where is Ethanol used?
Answer:
It is used in medicines such as tincture iodine, cough syrups, and many tonics.

Question 8.
What is vinegar? Mention one of its use.
Answer:
5 to 8% solution of acetic acid in water is called vinegar. It is used as a preservative in pickles. We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity

KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity.

Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity

KSEEB SSLC Class 10 Science Chapter 12 Intext Questions

Text Book Part I Page No. 94

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electric current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of electric current is ampere (A).
When I coulomb of electric charge flows through any cross. Section of a conductor in I second, the electric current flowing through it is said to be 1 ampere.
∵ 1 ampere = 1C/1s

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
The SI unit of electric charge is column, which is equivalent to the charge contained in nearly 6 × 1018 electrons.

Text Book Part I Page No. 96

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Voltmeter.

Question 2.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
One Volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 96 Q 2
1 V = 1 Jc-1

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
W = VQ
= 6 × 1 = 6 joules
Hence 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.

Text Book Part I Page No. 103

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of the conductor depends

  1. on its length
  2. on its area of cross section and
  3. on the nature of its material.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Resistance, \(\mathrm{R} \alpha \frac{1}{\mathrm{A}}\). The resistance of a conductor is inversely proportional to its area of cross-section. A thick wire has a greater area of cross-section whereas a thin wire has a smaller area of cross-section. Thus, thick wire has less resistance and a thin wire has more resistance therefore current will flow more easily through a thick wire.

Question 3.
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
As per ohm’s law, V = IR
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\) —— (i)
Potential difference is half
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 103 Q 3
∴ Current flowing is also half of ts former value.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
a) Resistivity of iron = 10.0 × 10-8
Resistivity of Mercury = 94.0 × 10-8
Resistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.
b) Resistivity of silver is less, hence it is a good conductor of heat.

Text Book Part I Page No. 107

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.
Answer:
Rs = R1 + R2 + R3
= 2V + 2V + 2V
= 6W
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 107 Q 1

Question 2.
Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 107 Q 2
Resistance, R = 5 + 8 + 12 = 25 Ω
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{25}=0.24 \mathrm{A}\)
As per ohm’s law =0.24A
V1 = IR
Potential difference
V1 = IR = 0.24 × 12 = 2.88 V.
reading in ammeter = 0.24A
reading in voltmeter = 2.88 v.

Text Book Part I Page No. 110

Question 1.
Judge the equivalent resistance when the following are connected in parallel –
(a) 1 Ω and 106 Ω,
(b) 1 Ω and 103 Ω, and 106 Ω.
Answer:
(a) If two resistances are connected in parallel
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 1
equivalent resistance = 1Ω.
(b) If three resistances 1Ω, 103Ω and 106Ω are connected in parallel,
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 1.1
equivalent resistance = 0.999Ω

Question 2.
An electric lamp of 100Ω, a toaster of resistance 50Ω and a water filter of resistance 500Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Resistance of Electric 1 amp R1 = 100Ω
Resistance of Tosser, R2 = 50Ω
Resistance of water filter, R3 = 50Ω
Potential difference, V = 220V
When these are connected in parallel,
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 2
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 2.1
7.04A of electricity is obtained by three appliances
Resistance of an electric iron connected to the same source that takes as much current as all three appliances
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 2.2
∴ Resistance of iron box = 31.25Ω
Electricity flowing through this = 7.04A.

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
(i) In parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.

(ii) In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In series circuit, appliances do not get the same voltage, as that of the power supply line.

(iii) In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much due to which the current from the power supply is low.

Question 4.
How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?
Answer:
In this diagram 2 resistors of resistance 3W and 3 W are connected in parallel.
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 4
If 2Ω and 2Ω are connected in series 2Ω + 2Ω = 4Ω
∴ Total resistance = 4Ω
If resistors are connected in series
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 4.1
∴ Total resistance = 1Ω

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, 24Ω?
Answer:
(a) If four resistors are connected in series then resistance
= 4 + 8 + 12 + 24
= 48Ω
(b) If these are connected in parallel
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 110 Q 5
∴ We can get lowest resistance.

Text Book Part I Page No. 112

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
In cord of an electric heater, as current flows these become hot arid glows but in case of electric heater this will not happen.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
As per Joule’s law H = VIt.
Here V = 50 V.
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity 112 Q 2

Question 3.
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
As per Joule’s law
H = VIt
H = IR = 5A × 20Ω = 100V
I = 5A, t = 30 sec.
∴ H = 100 × 5 × 30 J
= 15000 J = 1.5 × 104 J.

Text Book Part I Page No. 114

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
The rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power.
P = VI.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
P = VI
V = 220V, and I = 5A.
Power P = 220 × 5 = 1100 W
Power of the motor = P × t
P = 1100 W.
t = 2 Hr 2 × 60 × 60 W
= 7200 S
∴ Energy consumed, E = 1100 × 7200 J
= 7920000
= 7.92 × 106J.

KSEEB SSLC Class 10 Science Chapter 12 Textbook Exercises

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R` } \) is –
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) \(\frac { { V }^{ 2 } }{ R } \)
Answer:
(b) IR2

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity 1.6 10-8Ω m. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?
Answer:
Area of the copper wire A \(=\pi\left(\frac{\mathrm{D}}{2}\right)^{2}\)
diameter = 0.5 mm = 0.0005 m
Resistance, R = 10W.
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 6
Diameter of copper wire is doubled, then diameter = 2 × 0.5 = 1 mm = 0.001 m
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 6.1
∴ Length of wire = 122.7 m
Resistance of wire = 2.5 W

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes) 0.5 1.0 2.0 3.0 4.0
V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
In the following graph, voltage is taken along the x-axis and the current is taken along the y-axis. Different values are as follows:

V (volts) 1.6 3.4 6.7 10.2 13.2
I (amperes) 0.5 1.0 2.0 3.0 4.0

KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 16
Slope indicates Resistance
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 7.1
∴ Resistance of the resistor = 3.4Ω.

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Here I = 2.5 mA = 2.5 × 10-3 A.
∴ \(R=\frac{12}{2.5 \times 10^{-3}}\)
= 4.8 × 103
= 4.8 K cal.

Question 9.
A battery of 9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0. 4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ω resistor?
Answer:
A battery of 9V is connected in series with resistors 0.2Ω, 0.3Ω, 0.4Ω,
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Ω.
V = 9V
\(I=\frac{9}{13.4}=0.671 \mathrm{A}\)
∴ 12 V battery, 0.671 A current flows.

Question 10.
How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Let the resistors be x’
Resistance = 176Ω
As per ohm’s law,
V = IR
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 10
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 10.1
∴ Four resistors (in parallel) are required to carry 5A on a 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Answer:
If resistors are connected in series 6Ω + 6Ω + 6Ω =18Ω This ia not correct
When they are connected in parallel
\(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3\) This is also wrong,
i) When they are connected in parallel
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 11
Two 6Ω resistors are connected in parallel
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 11.1
If 3rd resistor of 6Ω and 3Ω are connected in series, it becomes 6Ω + 3Ω = 9Ω.
ii) When they are connected in series
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 11.2
Resistance = 6Ω + 6Ω
= 12Ω
If 3rd resistor 6Ω is connected to 12Ω in parallel
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 11.3
Total resistance = 4Ω.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Here V = 220V
I – 5A
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 12
∴ 110 lamps can be connected in parallel with each other across the two wires.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
(i) If coils are connected separately V = 220 V
Resistance R1 = 240
As per ohm’s law, V = IR
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 13
∴ If coils are connected separately 9.16A electricity flows in the coil.
(ii) If coils are connected in series
Resistance R2 = 24Ω + 24Ω = 48Ω
As per ohm’s law V = IR
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 13.1
∴ If coils are connected in series 4.58A electricity flows.

Question 14.
Compare the power used in the 20 resistors in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel with 12Ω and 2Ω resistors.
Answer:
(i) Potential Difference V = 6V
If 1Ω and 2Ω resistors are connected in series, then Resistance
R = 1 + 2 = 3W.
As per ohm’s law
\(\mathrm{I}=\frac{6}{3}=2 \mathrm{A}\)
P(I2)R = (2)2 × 2 = 8W.
(ii) Potential difference V = 4V
If 12Ω and 2Ω resistors are connected in parallel, voltage is equal
Voltage of resistance 2 W is 4 volts
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 14
∴ Power of 2Ω is 8 W.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Both lamps are connected in parallel potential difference = 220 V
Power = V × I.
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ex Q 15

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
250W TV set in used in 1 Hr, then its energy
= 250 × 3600 = 9 × 105
energy of Toaster = 1200 × 600
If it is used in 10 minutes, then its power = 1200 × 600 × 7.2 × 105 J
∴ Energy of 250 W TV set is used in 1 Hr is greater than 1200 W toaster used in 10 minutes.

Question 17.
An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours.
Calculate the rate at which heat is developed in the heater.
Answer:
P = I2R
R = 8Ω, I = 15A
P= (15)2 × 8 = 1800 J/s.
∴ Rate at which heat is developed in the heater
= 1800 J/s.

Question 18.
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Answer:
Tungstan is a strong and metal having high melting point. This will not melt in high temperature. Because of this tungsten is used almost exclusively for filament of electric lamps.

(b) Why are the conductors of electric heating devices, such as bread- toasters and electric irons, made of an alloy rather than a pure metal?
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. Because of this reason the conductors of electric heating devices, such as bread-toasters and electric irons made of an alloy rather than a pure metal.

(c) Why is the series arrangement not used-for domestic circuits?
Answer:
In case of series connection, when one component fails, the circuit it broken and none of the components work. But in case of parallel connection, circuit divides the current throughout the electrical gadgets. Because of this reason series arrangement is not used for domestic purposes.

(d) How does the resistance of a wire vary with its area of cross-section?
Answer:
Resistance is inversely proportional to its cross section. As area is increasing resistance is less. Thus resistance of wire is changing with area of cross section.

(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
Copper and Aluminium are good conductors of electricity and they have less resistance. Hence they are used in electricity transmission.

KSEEB SSLC Class 10 Science Chapter 12 Additional Questions and Answers

Question 1.
Draw a Neat diagram of Electric circuit for studying ohm’s law.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity Ad Q 1

Question 2.
What is the S.I. Unit of resistivity?
Answer:
ΩM

Question 3.
How is fuse wire made of?
Answer:
It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc.

Question 4.
Write the formula of Electric power.
Answer:
\(P\quad =\quad \frac { { V }^{ 2 } }{ R } \)

Question 5.
Give examples where we find heating of electric current.
Answer:
Electric heater, electric iron etc.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 1
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency ?
Solution:
(i) Here the maximum class interval is 35 – 45. This class interval has mode
l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.1
∴ Mode = 36.8 years

(ii) Mean (\(\overline{X}\)) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.2
Here, a = 30
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 1.3
= 30 + 5.37
= 35.37
∴ Mode = 36.8
∴ Mean = 35.37
Hence Age group 36.8 patients are admitted more in number.
Mean age group patients are 35.57

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 2
Determine the modal lifetimes of the components:
Solution:
Class interval which has mode = 60 – 80
Here, l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 2.1
∴ Mode = 65.63 Hours

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also. And the mean monthly expenditure :
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3
Solution:
(i) Mode is in class interval 1500 – 2000
Here, l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.1

(ii) Mean (\(\overline{X}\)): Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.2
Here, a = 2750
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 3.4
= 2750 – 87.50
= 2662.5
∴ Modal monthly expenditure = 1847.8
∴ Mean Monthly expenditure = Rs. 2662.5

Question 4.
The following distribution gives the state- wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4
Solution:
(i) Class interval which has mode 30 – 35
Here, l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4.1
∴ Mode = 30.6
(ii) Mean (\(\overline{X}\)) : Step Deviation Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 7
By Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 4.3
= 32.5 – 3.3
= 29.2
∴ Mode = 30.6
∴ Mean = 29.2
We conclude that more number of states has teacher-student ratio about 30.6.
Mean of Age ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 5
Find the mode of the data.
Solution:
Class interval which has mode is = 4000 – 5000
Maximum frequency is
Here, l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 5.1
= 4000 + 608.7
= 4608.7
∴ Mode = 4608.7 runs.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 6
Solution:
Class interval which has mode is 40- 50
Maximum frequency is 20.
Here, l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 Q 6.1
= 40 + 4.7
= 44.7
∴ Mode = 44.7 cars.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution

KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution.

Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution

KSEEB SSLC Class 10 Science Chapter 9 Intext Questions

Text Book Part II Page No. 53

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B have higher percentage hence, it is likely to have arisen earlier. In asexual reproduction, there would be only very minor differences generated due to small inaccuracies in DNA copying, so trait B, which exists in 60% of the same population may get inherited earlier while trait A which exists in 10% of the population may be originated late due to variations.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Variation is a process which occurs sometimes at cellular level of reproduction. It cause small changes in the genotype natural selection selects the individuals having useful variations which ensure their survival in the prevailing conditions of environment, variant individuals that can withstand or cope with prevailing environment will survive letter and will increase in number through reproduction.

Text Book Part II Page No. 57

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
The trait which appears in all the members of F1 generation and also in 75% numbers of F2 generation obtained by self fertilisation of F1 generation is dominant character.
The trait which does not appear in F1 generation but after selffertilisation of F1 generation, reappears in 25% of F2 generation is known as recessive.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel thought that if two different characteristics, rather than just one are bred with each other. What do the progeny of a tall plant with round seeds and a short plant with wrinkled seeds look like? They are all tall and have round seeds. Tallness and round seeds are thus dominant traits, But what happens when these F1 progeny are used to generate F2 progeny by self pollination? A Mendelian experiment will find that some F2 progeny are tall plants with round seeds and some were short plants with wrinkled seeds. However there would also be some F2 progeny that showed new combinations. Some of them would be tall, but have wrinkled seeds, while others would be short, but have round seeds, you can see as to how new combinations of traits are formed in F2 offspring when factors controlling for seed shape and seed colour recombine to form zygote leading to form F2 offspring. Thus the tall/short trait and the round seed/wrinkled seed trair are independently inherited.
KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution 57 Q 2

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No this information is not sufficient to determine which of the traits: blood group A or O is dominant. Blood group A can be genotypically AA or AO. Hence, the information is incomplete.

Question 4.
How is the sex of the child determined in human beings?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution 57 Q 4

Most human chromosomes have a maternal and a paternal copy, and we have 22 such pairs. But one pair, called the sex chromosomes, is odd in not always being a perfect pair. Women have a perfect pair of sex chromosomes, both called x. But men have a mismatched pair in which one is a normal-sized x, while the other is a short one called y. So women are xx, while men are xy. As fig. shows half the children will be boys and half will be girls. And children will inherit an x chromosome from their mother regardless of whether they are boys of girls. Thus the sex of the children will be determined by what they inherit from their father. A child who inherits an x chromosome, from her rather will be a girl, and one who inherits a y chromosome from him will be a boy.

Text Book Part II Page No. 60

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:

  1.  Natural selection,
  2.  Genetic drift.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Traits acquired physically, emotionally are not a change which affects the genotype of an individual so it does not get inherited for coming generations and also acquired trait involves change in non-reproductive tissues which cannot be passed on to germ cells or the Progeny. Therefore, these traits cannot be inherited.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small number of tigers indicates that tiger variants are having many challenges and not capable to adopt the existing environment and may extinct soon. The small number of members in a population of tigers may cause small number of variations, which are essential for the survival of the species. A deadly disease, or calamity may be fatal to alhthe tigers. Since, tigers are among the top consumers in our ecosystem hence, their presence in our surrounding is a must.

Text Book Part II Page No. 61

Question 1.
What factors could lead to the rise of a new species?
Answer:
Genetic drift and Natural selection are the two factors which lead to the rise of a new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self pollinating plant species? Why or why not?
Answer:
No, geographical isolation cannot prevent speciation in this case, since the plants are self-pollinating, which means that the pollens are transferred from the anther of one flower to the stigma of the same flower or of another flower of the same plant. Geographical isolation can prevent the transfer of pollens among different plants only and not pollination.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, because geographical isolation does not affect much in asexually reproducing organisms. Asexually reproducing organisms pass on the parent DNA to offsprings that leaves no chance of speciation. However, geographical isolation works as a major factor in cross pollinated species. As it would result in pollinated species and accumulation of variation in the two geographically separated population.

Text Book Part II Page No. 66

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
Birds and reptiles are great example of two close species. Feathers in some ancient reptiles, as fossils indicate, they were evolved to provide insulation in cold weather. However, they cannot fly with these feathers, later on birds adapted the feathers to flight. This means that birds are very closely related to reptiles, since dinosaurs were reptile.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat are similar in function i.e., flying. They look similar because of common use for flying, but their origins are different. Since, they perform similar function, they are analogous organs and not homologous.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Usually when organisms die, their bodies will decompose and be lost. But every once in a while, the body or at least some parts may be in an environment that does not let it decompose completely. If a dead insect gets caught in hot-mud, for example, it will not decompose quickly, and the mud will eventually harden and retain the impression of the body parts of the insect. All such preserved traces of living organisms are called fossils.
Fossils explain about the extinct species every existed.

Text Book Part II Page No. 68

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Because all humans are a single species.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:

Evolution cannot always be equated with progress or better body designs. Evolution simply creates more complex body designs. However, this does not mean that the simple body designs are inefficient. In fact, bacteria having a simple body design are still the most cosmopolitan organisms found on earth. They can survive in hot springs, deep sea and even freezing environment.

Therefore, bacteria, spiders, fish and chimpanzees are all different branches of evolution.

KSEEB SSLC Class 10 Science Chapter 9 Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW.

Question 2.
An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) all of the above.

Question 3.
In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
This information is not complete. We cannot say anything about whether the light eye colour trait is dominant or recessive only one generation is there.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:

Classification shows how clearly organisms are closely related with respect do evolution. As we know that each organism has descended from its ancestral type with some modification.

Classification involves grouping of organism based on similarities in internal and external structure or evolutionary history.

Two species are more closely related, if they have more characteristics in common. Different organisms would have common features if they are inherited from a common ancestor. And, if two species are more closely related, then it means they have a more recent ancestor. With subsequent generations, the variations make organisms more different than their ancestors. This discussion clearly proves that we classify organisms according to their resemblance which is similar to creating an evolutionary tree.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Homologous organs are those organs of different organisms which have the same basic structural design and origin but perform different functions. Example: the forelimbs of humans and the wings of birds look different externally but their skeletal structure is similar.

Analogus organs are those organs of different organism which have the different basic structural design and origin but have similar functions. Example: the wings of birds and bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Dog has 11 genes A….T. It acquires one chromosome from his father or mother. As per genetics Dog may be black or brown. They have 25% of BB and 50% Bb and 25% bb of genes. It is as follows.

B b
B BB Bb
b Bb bb

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:

The fossils are the remains of organisms that once existed on earth.

Fossil provide us evidence about

  • The organisms that lived long ago , their structure etc.
  • Evolutionary development of particular species i.e., line of their development with time and other environmental factors.
  • Connecting links between groups. For example, feathers present in reptiles means that birds are very closely related to reptiles.
  • Which organisms evolved earlier and which later.
  • Development of complex body designs from the simple body designs.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
J.B.S. Haldane, a British scientist (who became a citizen of India later), suggested in 1929 that life must have developed from the simple in organic molecules which were present on earth soon after it was formed. How did these organic molecules arise? An answer was suggested by the experiment conducted by Stanley L. Miller and Harold C Urey in 1953.

They assembled an atmosphere similar to that thought to exist on early earth (this had molecules like ammonia, methane) and Hydrogen sulphide, but no oxygen) over water. This was maintained at a temperature just below 100° c and sparks were passed through the mixture of gases to stimulate lighting. At the end of a week, 15% of the carbon (from methane) had been converted to simple compounds of carbon including amino acids which make up protein molecules. This is the evidence we have for the life from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexusal reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Sexual reproduction causes more viable variations due to the following reasons:

Two different types of gametes meet to form new individuals which have better possibilities of combinations of traits. Due to the presence of multiple traits, error in copying of DNA are highly significant. At the time of gamete formation, random aggregation of paternal and maternal chromosome occurs. Exchange of genetic material may occur between , homologous chromosomes during formation of gametes. Due to sexual reproduction over generation after generation and selection by nature created wide diversity. In case of asexual reproduction, only the very small changes due to inaccuracies in DNA copying pass on the progeny. Thus, offsprings of asexual reproduction are more or less genetically similar to their parents. So, it can be concluded that evolution in sexually reproducing organisms proceeds at a faster pace than in asexually reproducing organisms.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Each cell will have two copies of each chromosome, one each from the male and female parents. Every germ cell will take one chromosome from each pair and these may be of either maternal or paternal origin. When two germ cells combine, they will restore the normal number of chromosomes in the progeny, ensuring the stability of the DNA of the species. Hence there is equal genetic contribution of male and female parents ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, we agree with the statement that only variations that confer an advantage to an individual organism will survive in a population. Genetic variations are part of natural selection. Genetic variations are one form of variation. Selection is done with respect to the environment. Every variation do not have a survival in the environment in which they exist. The chances of surviving depend upon adaptation by surrounding and the nature of variations. Selection of variants by environmental factors forms the basis for revolutionary process.

KSEEB SSLC Class 10 Science Chapter 9 Additional Questions and Answers

Question 1.
How do we know how old the fossils are?
Answer:
There are two components to this estimation one is relative. If we dig into the earth and start finding fossils, it is reasonable to suppose that the fossils we find closer to the surface are more recent than the fossils we find in deeper layers. The second way of dating fossils is by detecting the ratios of different isotopes of the same element in the fossil material. It would be interesting to find out exactly how this method works!

Question 2.
What is Evolution?
Answer:
Evolution is simply the generation of diversity and the shaping of diversity by environmental selection.

Question 3.
What is scientific name of man?
Answer:
Home sapients.

Question 4.
What does evolution of human beings indicate?
Answer:
It indicates that all of us belong to a single species that evolved in Africa and spread across the world in stage.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1

(Unless stated otherwise, take π \(=\frac{22}{7}\))

Question 1.
2 cubes each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
If measurement of each side of each cube be l cm, then
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 1
(l)3 \(=\sqrt[3]{64}\)
∴ l = 4 cm.
Length of cubes if both joined together,
Length, l = 4 + 4 = 8 cm.
breadth, b = 4 cm.
height, h = 4 cm.
Surface area of the cuboid,
= 2lb + 2lh + 2bh
= 2(8 × 4) + 2(8 × 4) + 2(4 × 4)
= 2 × 32 + 2 × 32 + 2 × 16 = 64 + 64 + 32
= 160 sq.cm.

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
(i) Diameter of a hollo hemisphere = 14 cm.
∴ Radius of a hollow hemisphere, r
r = 7 cm.
Inner curved surface of a hemisphere = 2π2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 2.1
= 14 × 22
= 308 sq. cm.
(ii) Radius of hollow cylinder, r = 7 cm.
Height, h = 13 – 7 = 6 cm.
∴ Inner Surface area of the Cylinder,
= 2πrh
\(=2 \times \frac{22}{7} \times 7 \times 6\)
= 44 × 6
= 264 sq.cm.
∴ Inner surface area of the vessel = Area of hemisphere + Area of cylinder
= 308 + 264
= 572 sq. cm.

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
(i) Radius of base of circular toy,
r = 3.5 cm.
∴ Surface Area of hemisphere, = 2πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.1
= 77 sq.cm
(ii) Let slant height of cone is l cm. then
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.2
∴ Curved surface area of a cone = πrl
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 3.4
= 137.5 sq.cm.
∴ Total surface area of a toy = Surface area of hemisphere + Curved surface area of a cone
= 77.0 + 137.5
= 214.5 sq.cm.

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution:
(i) A cubical block of side 7 cm is surmounted by a hemisphere.
∴ Diameter of hemisphere is = 7 cm.
Radius of hemisphere, r = 3.5 cm.
∴ Surface area of hemisphere = Curved surface area – Area of base of hemisphere
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 4
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 4.1
= 38.5
(ii) The curved surface area of square,
=6 × l2
= 6 × (7)2
= 6 × 49
= 294 sq. cm.
∴ Total surface area of cubical block, = Curved surface area of cubical block + Surface area of hemisphere.
= 294 + 38.5
=332.5sq.cm.

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the cube. Determine the surface area of the remaining solid.
Solution:
Let, Length of each edge of cubical block = Diameter of hemisphere = l
∴ Radius of hemisphere, \(r=\frac{l}{2} unit\).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 5
Total surface area of newly formed cube = Curved surface area of cube – Upper part of Cube + Area of hemisphere which is depressed
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 5.1

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Figure given below). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6.1
(i) Surface area of cylindrical capsule, = 2πrh
\(=2 \pi \times\left(\frac{5}{2}\right) \times 9\)
= 45π mm2,
(ii) Surface area of both hemispheres,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 6.2
= 25π mm2
∴ Total surface area of capsule, = 45π + 25π mm2
= 70π mm2
\(=70 \times \frac{22}{7}\)
= 220 mm2

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1m and 4m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m2. (note that the base of the tent will not be covered with canvas.)
Solution:
(i) Radius of base of cylinder, r= 2 mm.
Height of base of cylinder, h= 2.1 m.
∴ Curved Surface area of cylinder = 2πrh
= 2π × 2 × (2.1)
\(=4 \times \frac{22}{7} \times 2.1\)
= 26.4 m2.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 7
(ii) Radius of conical tent,
r = 2 m.
length, l = 2.8 m.
∴ Curved Surface area of cylinder = πrl
\(=\frac{22}{7} \times 2 \times 2.8\)
= 17.6 m2.
∴ Total area of the tent = 26.4 + 17.6
= 44m2
Cost of 1 aq.m. of canvas is Rs. 500.
Cost of 44 sq. m. canvas … ??
500 × 44
= Rs. 22,000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Height of the cylinder, h = 2.4 m.
Diameter of the cylinder, d = 1.4 m.
∴ Radius, r = 0.7 m.
(i) Outer Curved surface area of cylinder = 2πrh
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8
\(=2 \times \frac{22}{7} \times(0.7) \times 2.4\)
= 44. × 0.24
= 10.56 cm2.
(ii) Area of base of = πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8.1
= 1.54 cm2.
(iii) Inner surface area of cylinder, = πrl
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 8.2
= 5.5 cm2
∴ Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm2.
∴ Nearest value is 18 sq.cm.

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm. and its base is of radius 3.5 cm. find the total surface area of the article.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9
Solution:
Radius of base of cylinder, r= 3.5 cm
Height, h = 10 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9.1
Total area of article, = Curved Surface area of Cylinder + 2 × Area of Hemisphere
= 2πrh + 2 × 2πr2
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 Q 9.2
= 374 sq.cm.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.1, drop a comment below and we will get back to you at the earliest.

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