Subject

Physics

Class

CBSE Class 12

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CBSE Physics 2012 Exam Questions

Short Answer Type

11.

Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’.

How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’? 

When a capacitor is charged by a battery, work is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy.

Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential difference between capacitor plates =0. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases.

 

Potential difference between its plates, V= q/C

Work done to give an infinitesimal charge dq to the capacitor is given by, 

dW space equals space straight V space dq space equals space straight q over straight C dq
space space space space space equals 1 over straight C open square brackets straight q squared over 2 close square brackets subscript 0 superscript straight Q space

space space space space space equals space fraction numerator straight Q squared over denominator 2 straight C end fraction 

If V is the final potential difference between capacitor plates, then Q = CV

therefore space straight W space equals space fraction numerator left parenthesis CV right parenthesis squared over denominator 2 straight C end fraction space equals 1 half CV squared space equals space 1 half QV

Work is stored in the form of electrostatic potential energy.

Electrostatic potential energy, U = fraction numerator straight Q squared over denominator 2 straight C end fraction space equals space 1 half C V squared space equals space 1 half Q V

When battery is disconnected,

i) Energy stored will decrease.

Energy becomes, U = fraction numerator straight Q subscript straight o squared over denominator 2 straight C end fraction space equals space fraction numerator Q subscript o squared over denominator 2 K C subscript o end fraction space equals space U subscript o over K

So, energy is reduced to 1/K times its initial energy.

 

i) In the presence of dielectric, electric field becomes, E = straight E subscript straight o over straight K

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12.

How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? 


Angular separation is given by,

straight theta space equals space straight beta over straight D space equals space fraction numerator Dλ divided by straight d over denominator straight D end fraction equals straight lambda over straight d

Angular separation would remain same when the distance of separation between the slit and the screen is doubled.

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13.

Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope


 

 

 

Advantages of reflecting telescope over refracting telescope:

(i) It is free from chromatic aberration.

(ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.

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14.

For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? 


As per Snell’s law we have,

straight n space equals space fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space straight c over straight v 
For given i, v proportional tosin r 
r is minimum in medium A, so velocity of light is minimum in medium A.




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15.

A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.


Length of the wire remains same,

straight N subscript 1 space straight x space 2 πR space equals space straight N subscript 2 space straight x space space 2 straight pi space straight R over 2

therefore space straight N subscript 2 space equals space 2 space straight N  

Magnetic moment of a coil, m = NAI

For the coil of radius R, magnetic moment, 

straight m subscript 1 space equals space straight N subscript 1 IA subscript 1 space equals space straight N subscript 1 Iπ space straight R squared
Magnetic moment for coil of radius R/2,

straight m subscript 2 space equals space straight N subscript 2 space IA subscript 2 space equals space fraction numerator 2 straight N subscript 1 IπR squared over denominator 4 end fraction equals fraction numerator straight N subscript 1 IπR squared over denominator 2 end fraction

Therefore, 

straight m subscript 2 over straight m subscript 1 space equals space 1 colon thin space 2

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16.

An object AB is kept in front of a concave mirror as shown in the figure.

 

(i) Complete the ray diagram showing the image formation of the object.

(ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black?        

(i) Image formed will be inverted diminished between C and F.

 

ii) When the lower half of the mirror’s reflecting surface is painted black, the position of the image and its intensity will get reduced.

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17.

Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.            


The average time elapsed between two successive collisions is known as the relaxation time of free electrons drifting in a conductor.



Relation between straight tau and vd is given by, 

                    

Consider a conductor of length ‘l’, area of cross-section A and current density n.

Current flowing through the conductor is given by, 

         

Electric field applied across the ends is given by, E = V/l

So current flowing through the conductor becomes,

                       

Then, 

straight V over straight I space equals space fraction numerator ml over denominator ne squared straight tau space straight A end fraction

Using ohm's law, we get

 


Therefore, 

                                            straight rho space equals space fraction numerator straight m over denominator ne squared straight tau end fraction

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18.

A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?


As per De-broglies formula, 


straight lambda space equals space fraction numerator straight h over denominator square root of 2 mE subscript straight k end root end fraction 

Kinetic energy of proton is equal to kinetic energy of proton.

Since, mass of proton > mass of electron,

This implies, 

straight lambda subscript straight e space greater than space straight lambda subscript straight p
That is, wavelength of electron is greater than the wavelength of proton.

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19.

Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E? 


 


We have,


straight R subscript BCD space equals space 5 space straight capital omega space plus space 10 space straight capital omega space equals space 15 space straight capital omega

Effective resistance between B and E is,





On applying Kirchoff's law, we have





Hence,

  

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20.

A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.


Suppose a resistance R, inductance L and capacitance C connected in series.

An alternating source of voltage V = Vo sin straight omegat is applied across it. Since all the components are connected in series, the current flowing through all is same.

Voltage across resistance R is VR, voltage across inductance L is VL and voltage across capacitance C is VC.

VR and (VC -VL) are mutually perpendicular and the phase difference between them is 90°.

 

From the figure above, we have 


and






The phase difference between current and voltage is given by, 

                              



From the graph, we can see that with increase in frequency, current first increases and then decreases. At resonant frequency, current amplitude is maximum.

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