Subject

Physics

Class

CBSE Class 12

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CBSE Physics 2013 Exam Questions

Short Answer Type

11.

The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. 



The pair of curves (1, 3) and (2, 4) corresponds to different materials but having same intensity of incident radiation. The value of stopping potential for the curves (1, 2) and (3, 4) is same implying that they correspond to similar materials. 

1332 Views

12.

In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (i) the brightness of the lamp and (ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is decreased? Justify your answer.    


    


The given figure is a Common Emitter configuration of an npn transistor. Input circuit is forward biased and collector circuit is reverse biased.

i)

If the value of the resistance R is reduced, the current in the forward biased input circuit increases. The emitter current IE and the collector current IC (= IE – IB) both increase. Hence, the brightness of the lamp increases.

ii) 

Due to increase in IC, the potential drop across lamp L increases and hence the voltmeter reading V increases.

1356 Views

13.

(a) Write the necessary conditions for the phenomenon of total internal reflections to occur.

(b) Write the relation between the refractive index and critical angle for a given pair of optical media.

a) Conditions for total internal reflection to take place:

i) Light ray must travel from denser medium into rarer medium.

ii) The angle of incidence in denser medium must be greater than the critical angle for given pair of optical media. 

b) Relation between refractive index and critical angle for a given pair is,

    As per Snell’s law, 

fraction numerator sini subscript straight c over denominator sin 90 to the power of straight o end fraction equals n subscript 21 space equals space n subscript 1 over n subscript 2 italic space

therefore space space space space space space space space sin i subscript c space equals space n subscript italic 1 over n subscript italic 2 

1187 Views

14.

Block diagram of a receiver is shown in the figure:



(a) Identify ‘X’ and ‘Y’.

(b) Write their functions.


a) X represents Intermediate frequency stage and Y represents amplifier.

b) IF stage changes the electromagnetic wave of high frequency to a lower frequency for further detection in detector.

Function of amplifier is to enhance the power of the signals upto a required level.

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15.

A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it. 


As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.

Expression for Induced emf in a Rotating Rod

Consider a metallic rod OA of length l, which is rotating with angular velocity  in a uniform magnetic field B, the plane of rotation being perpendicular to the magnetic field. A rod may be supposed to be formed of a large number of small elements. Consider a small element of length dx at a distance x from centre. 

If v is the linear velocity of this element, then area swept by the element per second = v.dx 

Emf induced across the rod , dϵ space equals space d epsilon equals B fraction numerator d A over denominator d t end fraction equals space B space v space d x
But, v = straight x space straight omega

d epsilon space equals space B space x space omega space d x

Therefore,

Emf induced across the rod is given by, 

Error converting from MathML to accessible text. 

Power dissipated, when circuit is closed is, P = epsilon squared over R equals fraction numerator B squared omega l squared over denominator 4 R end fraction

4576 Views

16.

In a series LCR circuit connected to an ac source of variable frequency and voltage straight nu space equals space straight nu subscript straight m sinωt draw a plot showing the variation of current (I) with angular frequency (straight omega) for two different values of resistance R1 and R2 (R1>R2). Write the condition under which the phenomenon of resonance occurs. For which values of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance. 


At certain frequency straight omega space comma space the flow of current through the series combination,

straight I subscript straight m space equals space fraction numerator straight v subscript straight m over denominator square root of straight R squared plus left parenthesis straight chi subscript straight L minus straight chi subscript straight C right parenthesis squared end root end fraction

When comma space space straight chi subscript straight L space space equals space ωL space and space straight chi subscript straight c space equals space 1 over ωC

Condition of resonance – If system (LCR) of natural frequency w0 is driven by an energy source at a frequency w, the amplitude of the current flow increases, however the amplitude of the current rises to its maximum value, if frequency of the energy source becomes exactly equal to the natural frequency.

For resistance R2 < R1, series LCR shows a sharp resonance.

Q-factor - The ratio of reactance (either inductive or capacitive) at natural frequency to the resistance of the current is called Q - factor.

                                         straight Q space equals space straight chi subscript straight L over straight R equals space fraction numerator straight omega subscript straight o straight L over denominator straight R end fraction

Significance of Q factor is given by,

(i) If resistance R is low or inductance L is large then Q – factor is large and the circuit is more selective.

(ii) If resonance is less sharp, tuning of the circuit will not be good.

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17.

Write the relationship between angle of incidence ‘i’, prism ‘A’ and angle of minimum deviation for a triangular prism. 


The relationship between i, A and straight delta subscript straight m for a triangular prism is given by, 

                                           straight i space equals fraction numerator straight A plus space straight delta subscript straight m over denominator 2 end fraction

1749 Views

18.

Which of the following waves can be polarized (i) Heat waves (ii) Sound waves?


i) Heat waves can be polarized because they are transverse and electromagnetic in nature.

ii) Sound waves are longitudinal in nature and hence, cannot be polarized.

1437 Views

19.

Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect the optical signals. 

OR

Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?              


In any diode, an electric field ‘E’ exists across the junction from n-side to p-side, when light with energy h greater than energy gap Eg illuminates the junction. Then electron-hole pairs are generated due to absorption of photons, in or near the depletion region of the diode. Due to existing electric field, electrons and holes get separated. The free electrons are collected on n-side and holes are collected on p-side, giving rise to an emf.

Due to this generated emf, an electric current of straight muA order flows through the external resistance.



The figure above is the circuit diagram of photodiode.

Detection of optical signals:

Photodiode can be used to detect optical signals if we observe the change in current with change in light intensity when a reverse bias is applied.

                                                         OR

Important considerations in the fabrication of LED:

i) LED is heavily doped p-n junction.

ii) Reverse breakdown voltage of LED’s are very low, typically around 5 V.

The order of band gap of an LED to emit light in the visible range is about 3 eV to 1.8 eV.

1948 Views

20.

Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulate wave is obtained when a modulating signal is superimposed on a carrier wave.    


Three factors for the need of modulating a message signal:

a) Size of the antenna or aerial should be straight lambda/4.

b) Effective power radiated by an antenna should be close to Error converting from MathML to accessible text.
c) Interference of signals from different transmitters. Inorder to avoid the interference of the signals, high frequency is required which can be acquired by modulation. 

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