Subject

Physics

Class

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

11. Write three characteristic features in photoelectric effect that cannot be explained on the basis of wave theory of light, but can be explained only using Einstein's equation.

The three characteristics of photoelectric effect, which cannot be explained on the basis of wave theory of light are: 

1. For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light. 

2. For a given metal, a certain minimum frequency of the incident radiation below which no emission of photoelectrons take place. The is the threshold frequency. 

3. Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light and is dependent only upon the frequency or wavelength of the incident light. 

4071 Views

12.

Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.


(i) Attenuation: Attenuation is the process of loss in the strength of signal while propagating through a medium. 

(ii) Demodulation: Demodulation is the process of retrieval of information from the carrier wave at the receiver end. It is the reverse process of modulation.
1868 Views

13. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also.

(b) Using mirror formula, explain why does a convex mirror always produce a virtual image.


a) 

Given, 

Radius of curvature, R = 20 cm

So, focal length, f = R/2 = - 10 cm

Since the image obtained is real, therefore magnification of the image, m = -2 

Now, using the formula, 


space space space space straight m space equals space fraction numerator negative straight v over denominator straight u end fraction
rightwards double arrow space minus 2 space equals space fraction numerator negative straight v over denominator straight u end fraction space
rightwards double arrow space straight v space equals space 2 straight u space

Using space the space mirror space formula comma space

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

space space space space space equals space fraction numerator 1 over denominator 2 straight u end fraction space plus space space 1 over straight u space equals space fraction numerator 3 over denominator 2 straight u end fraction
rightwards double arrow space straight u space equals space begin inline style 3 over 2 end style space straight f space
space space space space space space space space space equals space 3 over 2 space straight x space left parenthesis negative 10 right parenthesis space
space space space space space space space space space equals space minus 15 space cm space

Therefore comma space

straight v space equals space space 2 straight u space equals space minus 30 space cm

Therefore, the distance of the object is 15 cm in front of the mirror and the position of the image is  30 cm, formed in front of the mirror. 

b) 

For a convex mirror, 

Focal length, f > 0 
Position of the object, u < 0

Using mirror formula, we have

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

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

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

therefore space straight v thin space greater than space 0 space

That is, the image formed by a convex lens is always behind the mirror and hence is virtual. 

2469 Views

14.

(a) Explain any two factors that justify the need of modulating a low-frequency signal.
 
(b) Write two advantages of frequency modulation over amplitude modulation.


Modulating low-frequency signals are required because: 

i) Size of the antenna: Antenna is required for both transmission and reception of the signal whose size (at least straight lambda over 4) should be comparable to the wavelength of the signal so that time variation of the signal can be properly sensed by the antenna. For a low-frequency signal, the wavelength is large. Therefore, a large antenna of such a huge height is practically impossible to construct and operate.

Hence, the need for high-frequency transmission arises. 

ii) Overlapping signals from different transmitters: When an information is transmitted using different transmitters, the signals get overlapped and the information is lost. Hence, high-frequency transmission is required. Each band should be allocated to each frequency range in order to avoid mixing of signals. 

iii) Higher energy: High-frequency signals have high energy and therefore, even after loss due to attenuation, signals can be carried over longer distances. 

b) 

Advantage of frequency modulation over amplitude modulation are: 

i) FM is more efficient than AM because the amplitude of an FM wave is constant, irrespective of its modulation index. Thus, the transmitted power is constant. Also, in AM the transmitted power goes waste in the transmitted carrier.  

ii) FM reception is quite immune to noise in comparison to AM reception. Noise is a form of amplitude variation in the transmitted signal. Thus, using amplitude limiters in the FM receivers, the noise is eliminated.   

1576 Views

15. (i) Write the functions of three segments of a transistor.

(ii) Draw a circuit diagram for studying the input and output characteristics of a n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.

i) The function of three segments of a transistor are: 
 
Emitter: Emits the majority charge carriers

Collector: Collects the majority charge carriers

Base: Base provides the interaction between the collector and the base

ii) An n-p-n transistor is as shown below: 


Input characteristics: The variation between the base current and the base-emitter voltage is obtained. Input characteristics is used to find input dynamic resistance of the transistor as it is represented by the slope. 

Output characteristics: A graph representing the variation of the collector current and emitter voltage is obtained, keeping base current fixed. The slope fo the output characteristic graph gives us the output dynamic resistance.

3044 Views

16. Two long, straight, parallel conductors carry steady currents, I1 and I2 , separated by a distance . If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other? Obtain the expression for this force. Hence, define one ampere.

                                       
Magnetic field induction at some point P on wire 2 due to current I1 passing through wire 1 is given by, 
                                       straight B subscript 1 space equals space fraction numerator straight mu subscript straight o space 2 space straight I subscript 1 over denominator 4 space straight pi space straight d end fraction 

Magnetic field is produced by wire 1 and current carrying wire 2 lies in magnetic field B1.
 The unit length of wire 2 will experience a force, given by 

                              F= B1 I2 x 1 = fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator 2 space I subscript 1 space I subscript 2 over denominator d end fraction

According to Fleming's left-hand rule, the force on wire 2 acts in the plane of paper perpendicular to wire 2, directed towards wire 1. Similarly, wire 1 also experiences the same force towards wire 2. Thus, both the conducting wires attract each other with the same force F.

One ampere can be defined as the amount of current flowing through two parallel conductors, which are in the same direction or opposite directions, placed at a distance of one metre in free space, and both the wires attract or repel each other with a force of 2 x 10-7 per metre of their lengths. 
4422 Views

17. Plot a graph showing variation of de Broglie wavelength λ versus fraction numerator 1 over denominator square root of straight V end fraction, where V is the is accelerating potential for two particles A and B, carrying the same charge but different masses m1 and m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? 

Given, 

Charge (q) is the same but the mass of both particles is different (m1 > m2).

The De-broglie wavelength given by, 


straight lambda space equals space fraction numerator straight h over denominator square root of 2 mqV end root end fraction
The slope of the graph of λ versus fraction numerator 1 over denominator square root of straight V end fraction  = fraction numerator straight h over denominator square root of 2 mqV end root end fraction

                                    
The slope of the smaller mass is larger; therefore, plat A in the above graph represents mass m2.

6669 Views

18. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments, each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.

                                               OR

Calculate the energy in the fusion reaction: 

straight H presubscript 1 presuperscript 2 space plus space straight H presubscript 1 presuperscript 2 space rightwards arrow He presubscript 2 presuperscript 3 space plus space straight n space comma

where space straight B. straight E. space of space straight H presubscript 1 presuperscript 2 space equals space 2.23 space MeV

space space space space space space space space space space space space straight B. straight E. space of space He presubscript 2 presuperscript 3 space equals space 7.73 space MeV

The B.E. of the nucleus of mass number 240, B1 = 7.6 x 240 = 1824 MeV

The B.E of each product nucleus, B2 = 8.5 x 120 - 1020 MeV

Then, the energy released as the nucleus breaks is given by, 

E = 2B2 - B1 = 2 x 1020 - 1824 = 216 MeV 

                                                 OR

Given: 

B.E of straight H presubscript 1 presuperscript 2 comma space straight E subscript 1 space equals space 2.23 space MeV
B.E of He presubscript 2 presuperscript 3 comma space straight E subscript 2 space equals space 7.73 space MeV
Energy in the fusion reaction is given by, 


increment straight E space equals space straight E subscript 2 space minus space 2 straight E subscript 1 space equals space 7.73 space minus space left parenthesis 2 space straight x space 2.23 right parenthesis space equals space 3.27 space MeV

3007 Views

19. How are em waves produced by oscillating charges?
 
Draw a sketch of linearly polarised em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.

                                                      OR

Write Maxwell's generalisation of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is I = straight epsilon subscript straight o space dϕ subscript straight E over dt, where straight ϕ subscript straight E is the electric flux produced during charging of the capacitor plates. 

A moving charge produces both electric and magnetic fields, and an oscillating charge produces oscillating magnetic and electric fields. These oscillating electric and magnetic fields with respect to space and time produce electromagnetic waves. 

The propagation of electromagnetic waves can be shown as:




                                                  OR


Maxwell's generalization of ampere's circuital law given by,


contour integral space B with rightwards harpoon with barb upwards on top. space space stack d l with rightwards harpoon with barb upwards on top space equals space mu subscript o space left parenthesis I space plus space I subscript D right parenthesis thin space equals space straight mu subscript straight o space left parenthesis space straight I space plus space straight epsilon subscript straight o space dφ over dt right parenthesis
Consider that a parallel capacitor C is charging in a circuit. 

The magnitude of electric field between the two plates will be, 

straight E space equals space fraction numerator straight q over denominator straight epsilon subscript straight o straight A end fraction , is perpendicular to the surface of the plate. 

Electric flux through the surface will be, 

straight ϕ subscript straight E space equals space straight E with rightwards harpoon with barb upwards on top. space straight A with rightwards harpoon with barb upwards on top space

space space space space space equals space EA space cos space straight theta

space space space space space equals space fraction numerator straight q over denominator straight epsilon subscript straight o space straight A end fraction space straight x space straight A space

space space space space space equals space straight q over straight epsilon subscript straight o

rightwards double arrow space dϕ subscript straight E over dt space equals space fraction numerator straight d space open parentheses begin display style straight q over straight epsilon subscript straight o end style close parentheses over denominator dt end fraction

rightwards double arrow space dq over dt space equals space straight epsilon subscript straight o space dϕ subscript straight E over dt
That space is comma space dq over dt space is space the space rate space of space change space with space time. space

rightwards double arrow space straight I thin space equals space straight epsilon subscript straight o space dϕ subscript straight E over dt

4481 Views

20. State Brewster's law. 
The value of Brewster's angle for a transparent medium is different for light of different colours. Give reason.

Brewster discovered a relation between polarising angle, ip and the refrcative index n of the transparent material with respect to the surrounding medium.

The law is given by, 

                                                   straight mu space equals space tan space straight i subscript straight p

When a light is incident on a transparent surface at the polarising angle, the reflected and the refracted rays are perpendicular to each other. 

The lights of different colours have different wavelength hence, there are different values of refractive index for a medium. That is why these lights have different values of brewster's angles.
3401 Views