## Question Bank in Physics Class XII

The collection will also be helpful for students of other syllabuses.

The file consists of syllabus, key points, collection of very short Answer (1 mark), Short answer type question – solved (2 marks), short answers (3 marks), Long answers (5 marks), Solved numericals and 3 sample papers.

The contents are arranged chapter-wise. Any student will find this a boon for easy preparation and to score better marks in Physics.

## Solutions to NCERT Physics Class 12 (Ray Optics)

1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:Size of the candle, h= 2.5 cmImage size = h’Object distance, u= −27 cmRadius of curvature of the concave mirror, R= −36 cmFocal length of the concave mirror, f=R/2 = -18 cm

Image distance = v

The image distance can be obtained using the mirror formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$=\frac{1}{-18}-\frac{1}{-27}=\frac{-3+2}{54}=-\frac{1}{54}$

Therefore, v=-54cm

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as: $m=\frac{h'}{h}=-\frac{v}{u}$

Therefore, $h'=-\frac{v}{u}\times&space;h&space;=&space;-\frac{-54}{-27}\times&space;2.5&space;=&space;-&space;5&space;cm$

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
4. Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)].
5. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
8. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
9. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
10. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,
13. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
14. (a)A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m, and the radius of lunar orbit is 3.8 × 108 m.

## Quick Revision for Class X Physics SA1

CHAPTER -12 ELECTRICITY

GIST OF THE LESSON

1. Positive and negative charges: The charge acquired by a glass rod when rubbed with silk is called positive charge and the charge acquired by an ebonite rod when rubbed with wool is called negative charge.
2. Coulomb: It is the S.I. unit of charge. One coulomb is defined as that amount of charge which repels an equal and similar charge with a force of 9 x 109 N when placed in vacuum at a distance of 1 meter from it. Charge on an electron = -1.6 x 10-19 coulomb.
3. Static and current electricities: Static electricity deals with the electric charges at rest while the current electricity deals with the electric charges in motion.
4. Conductor: A substance which allows passage of electric charges through it easily is called a ‘conductor’. A conductor offers very low resistance to the flow of current. For example copper, silver, aluminium etc.
5. Insulator: A substance that has infinitely high resistance does not allow electric current to flow through it. It is called an ‘insulator’. For example rubber, glass, plastic, ebonite etc.
6. Electric current: The flow of electric charges across a cross-section of a conductor constitutes an electric current. It is defined as the rate of flow of the electric charge through any section of a conductor.

Electric current = Charge/Time     or        I = Q/t

Electric current is a scalar quantity.

1. Ampere: It is the S.I. unit of current. If one coulomb of charge flows through any section of a conductor in one second, then current through it is said to be one ampere.                                                                                                                         1 ampere = 1 coulomb/1 second    or      1 A = 1C/1s = 1Cs-1                                                                                                                                                                                                              1 milliampere =    1 mA = 10-3 A                                                                                                                                                                                 1 microampere = 1µA = 10-6 A
2. Electric circuit: The closed path along which electric current flows is called an ‘electric circuit’.
3. Conventional current: Conventionally, the direction of motion of positive charges is taken as the direction of current. The direction of conventional current is opposite to that of the negatively charged electrons.
4. Electric field: It is the region around a charged body within which its influence can be experienced. Continue reading “Quick Revision for Class X Physics SA1”

## The problem of a rock thrown vertically up

A rock is thrown vertically upward from the ground with an initial speed 15m/s.

a. how high does it go
b. how much time is required for the rock to reach its maximum height?
c. what is the rock’s height at t=2.00s?

(Posted by Merhawi)

(a)

u=15m/s

a=-10m/s2

v=0 m/s (at the max height)

S=?

v2-u2=2aS

S=v2-u2/2a

=225/20

=11.25 m

(b) From the above case

using v=u+at

t=v-u/a=1.5s

(c) Use S = ut + 1/2 at2

put t=2s, u = 15m/s, a=-10m/s2

S=30 – 20 = 10 m

(If you use g = 9.8 m/s2 The answers will be slightly different)