CBSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

 Multiple Choice QuestionsShort Answer Type


The equivalent wavelength of a moving electron has the same value as that of a photon of energy 6 x 10–17 J. Calculate the momentum of the electron.

Momentum of electron is given by, p = straight h over straight lambda

Error converting from MathML to accessible text.



The line AB in the ray diagram represents a lens State whether the lens is convex or concave.


We can see that the lens slightly diverges the path of refracted light. Hence, the lens is concave in nature.



A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of angle of prism. Find the angle of deviation.


Calculate the speed of light in a medium whose critical angle is 45°. Does critical angle for a given pair of media depend on the wavelength of incident light? Give reason.          

Given, i = e = ¾ A,                              …(1)


A is the angle of prism.

Angle of deviation for a ray of light in a prism is given by,

straight delta space equals space straight i space plus space straight e space minus space straight A

Since space straight e space equals space straight i. space we space have

straight delta space equals space 2 straight i space minus space straight A

Therefore space from space eqn. space left parenthesis 1 space right parenthesis comma space we space have space

space space space space straight delta space equals space 2 space straight x space 3 over 4 space straight A space minus space straight A
rightwards double arrow space straight delta space equals space 1 half straight A

Prism is an equilateral triangle. 

That is A= 60o

rightwards double arrow space delta space equals space 1 half cross times 60 space equals space 30 degree space semicolon space is space the space required space
angle space of space minimum space deviation. space


Using the snell's law we have

mu space equals space fraction numerator 1 over denominator sin space C end fraction                                       ... (1)

where, C is the critical angle of the medium and straight mu is the refractive index of  the medium.

Also, mu space equals space c over nu                                     ... (2)

Now, from equations (1) and (2), we have 

straight c over straight nu equals straight space fraction numerator 1 over denominator sin straight space straight C end fraction

Error converting from MathML to accessible text.

This is the speed of the light in the medium. 

Refractive index of a medium is inversely proportional to the wavelength of light. So, critical angle also depends upon the wavelength of incident light.



How does one explain, using de Broglie hypothesis, Bohr's second postulate of quantization of orbital angular momentum? 

According to de-Broglie hypothesis, a stationary orbit is the one that contains an integral number of de-Broglie waves associated with the revolving electron.

Total distance covered by electron = Circumference of the orbit =  

For the permissible orbit,

2 pi r subscript n italic space italic equals italic space n italic space lambda                          ... (1)

Now, according to De-Broglie wavelength,

lambda equals fraction numerator h over denominator m v end fraction

Now, putting this in equation (1), we have

 2 pi r subscript n = nh over mv 
rightwards double arrow mvn rn =  fraction numerator nh over denominator 2 straight pi end fraction; which is the required Bohr’s second postulate of quantization of orbital angular momentum.



Write two important considerations used while fabricating a Zener diode. Explain, with the help of a circuit diagram, the principle and working of a Zener diode as voltage regular. 

Important considerations while fabricating Zener diode are: 

i) Heavily doped p and n junctions are to be used for fabricating Zener diode.

ii) The breakdown voltage of the material must be noted down to avoid destruction of the device.

Principle: A properly doped p–n ​junction that is supposed to work in a breakdown region is called a Zener diode. Such a junction when reverse biased shows a sudden increase in current to a high value at a certain voltage known as the breakdown voltage or the Zener voltage.

Working: A Zener diode is connected across the fluctuating voltage source through a resistance R. A constant voltage supply is maintained across the load resistance RL.

When the input voltage increases, the resistance of the Zener diode decreases and hence the current through the diode increases. Thus, a large voltage drop is seen across the resistance R and the output voltage across RL remains at a constant value. When the input voltage decreases, the resistance of the Zener diode increases and hence the current through the diode decreases. Only a small voltage drop takes place now across the resistance R and the output voltage at RL remains constant. Thus, we get a constant voltage in spite of the fluctuating input voltage. In this way, a Zener diode acts as a voltage regulator.



Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators.

How does the change in temperature affect the behaviour of these materials? Explain briefly. 

Energy band diagram is as given below:


i) The valence band is completely filled and the conduction band can have two possibilities—either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other.

ii) On applying an even small electric field, conductors can conduct electricity.


i) For insulators, the energy gap between the conduction and valence bands is very large. Also, the conduction band is practically empty. 

ii) When an electric field is applied across such a solid, the electrons find it difficult to acquire such a large amount of energy to reach the conduction band. Thus, the conduction band continues to be empty. That is why no current flows through insulators.


i) The energy band structure of semiconductors is similar to that of insulators, but in their case, the size of forbidden energy gap is much smaller than that of the insulators.

ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it comparatively easier to shift to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators. 



(a) Write the necessary conditions to obtain sustained interference fringes.

(b) In Young's double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve?

(c) What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?

a) Necessary conditions for sustained interference fringes are:

i)     Two sources must be coherent. They should emit continuous light waves of same wavelength or frequency.

ii)    Two sources of light must be narrow.

iii)  Sources of light must be monochromatic.

b) Fringe width is given by:

                                       straight beta space equals space λD over straight d
where, D is the distance between the slit and the screen.

Fringe width and distance of screen from the slit is a linear graph.

The slope of the curve gives is  straight lambda over straight d
That is, when fringe width varies linearly with distance of screen from the slits, the ratio of wavelength to distance between the slits remain constant. That is why it is advised to take wavelengths of incident light nearly equal to the width of the slit.

c) If the distance between the slits is reduced by keeping other parameters same. Then, fringe width would become broader.



What is ground wave communication? Explain why this mode cannot be used for long distance communication using high frequencies.       

Ground waves travel along the surface of the earth. In ground wave propagation, a large portion of wave energy is in space near the surface of the Earth. The propagation of the wave is guided along the Earth's surface and follows the curvature of the Earth.

High frequency wave propagation is not possible through ground waves for long-distance communication because while progressing, ground waves induce current in the ground and bend round the corner of the objects on the Earth due to which the energy of the ground waves of high frequency is almost absorbed by the surface of the Earth after travelling a small distance. Such signals are also absorbed by the obstacles (like mountains, tall buildings and trees) between the transmitter and the receiver. This loss in power of ground waves increases with the increase in frequency. Thus, ground wave communication is not suited for high frequency.



(a) Give two reasons to explain why reflecting telescopes are preferred over refracting type. 

(b) Use mirror equation to show that convex mirror always produces a virtual image independent of the location of the object.         

a) Reflecting telescopes are preferred over refracting because of the following reasons:

i) There is no chromatic aberration as the objective is a mirror

ii) Spherical aberration is reduced in case of reflecting telescope by using mirror objective.

b) Focal length is always positive for convex mirror.

Consider an object placed on the left side of the mirror. That is u < 0.

Now, using the mirror formula,

1 over straight f space equals space 1 over v space plus space 1 over u

rightwards double arrow space 1 over v space equals space 1 over f minus 1 over u

Focal length is positive and object distance is negative.


1 over straight v greater than 0
rightwards double arrow space v space less than space 0 space

Hence, a virtual image is always formed at the back side of a mirror. That is, image formed by a convex mirror is virtual in nature.



(a) What are the three basic units in communication systems? Write briefly the function of each of these.

(b) Write any three applications of the Internet used in communication systems. 

a) A communication system consists of three basic units:

(i) Transmitter: This unit is used for transmitting the information after modifying it to a suitable form. It basically consists of a transducer that converts signal in any physical form to electrical signal for transmission. After that, the signal is modulated to transmit over long distances.

(ii) Communication channel: This unit carries the modulated signal from the transmitter to the receiver. Transmission lines act as a communication channel in case of telephony, whereas the free space serves the purpose of communication channel in case of the radio communication.

(iii) Receiver: This unit consists of an antenna, which receives the signal, followed by a demodulator, an amplifier and a transducer. The demodulator demodulates the modulated signal, the amplifier boosts up its intensity and the transducer converts it back again from electrical form to the needed physical form.

b) Application of internet used in communication system is:

i) E-banking: Financial transactions can be made through this mode.

ii) Internet surfing: Navigation over World Wide Web (www) from one webpage/website to another is called Internet surfing. It is an interesting way of searching and viewing information on any topic of interest.

iii) E-commerce: This is a mode of electronic commerce by which consumers can buy the products or the services over internet.