Subject

Physics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

11. Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of the depletion region and barrier potential in a p-n junction.  

Diffusion and Drift are the two processes that take place in the formation of a p-n junction.

In an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a P -type semiconductor, the concentration of holes is more than the concentration of electrons. During the formation of p-n junction.

Due to the diffusion of electrons and holes across the junction a region of
(immobile) positive charge is created on the n-side and a region of(immobile) negative charge is created on the p-side, near the junction; this is called depletion region.

Barrier potential is formed due to loss of electrons from n-region and gain of electrons by p-region. Its polarity is such that it opposes the movement of charge carriers across the junction.

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12.

(i) Obtain the expression for the cyclotron frequency.
(ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.


When a charged particle (q) moves inside a magnetic field B with velocity v, it experiences a force F = q(v xB) when v is perpendicular to B, the force F on the charged acts as the centripetal force and makes it move along a circular path.
Let m be the mass of particle and r is radius  of circular path

straight q space left parenthesis straight v with rightwards arrow on top space straight x straight B with rightwards arrow on top right parenthesis space equals space mv squared over straight r
qvB space equals space mv squared over straight r
rightwards double arrow space straight r space equals space mv over Bq

Time period of the circular motion of a charged particle is given by

straight T space equals space fraction numerator 2 πr over denominator straight v end fraction
space equals fraction numerator 2 πmv over denominator straight v space Bq end fraction
rightwards double arrow space straight T space equals space fraction numerator 2 πm over denominator Bq end fraction
Angular space frequency space comma
straight omega space equals space fraction numerator 2 straight pi over denominator straight T end fraction
therefore space straight omega space equals space Bq over straight m
ii) No, The mass of the two particles, i.e deuteron and proton, is different.Since (cyclotron) frequency depends inversely on the mass, they cannot be accelerated by the same oscillator frequency.

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13.

Name the junction diode whose I-V characteristics are drawn below


Solar cell

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14.

Write two important limitations of Rutherford nuclear model of the atom.


1. According to Rutherford model, an electron orbiting around the nucleus continuously radiates energy due to the acceleration; hence the atom will not remain stable.
2. As electron spirals inwards; its angular velocity and frequency change continuously; therefore it will emit a continuous spectrum.

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15. Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain.

Magnifying power is defined as the angle subtended at the eye by the image
to the angle subtended (at the unaided eye) by the object.
straight m space equals space straight m subscript straight o space straight x space straight m subscript straight e space equals space straight L over straight f subscript straight o space straight x straight D over straight f subscript straight e

To increase the magnifying power both the objective and eyepiece must have short focal lengths such as straight m space space equals space straight L over straight f subscript straight o space straight x straight D over straight f subscript straight e 

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16.

Define the distance of closest approach. A α-particle of kinetic energy 'K' is bombarded on a thin gold foil. The distance of the closest approach is 'r'. What will be the distance of closest approach for a α-particle of double the kinetic energy?   


It is the distance of charged particle from the centre of the nucleus, at
which the whole of the initial kinetic energy of the (far off) charged
particle gets converted into the electric potential energy of the system.
Distance of closest approach (rc) is given by

straight r subscript straight c space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction. fraction numerator 2 space Ze squared over denominator straight K end fraction
(rc) is inversely proportional to K. Thus,
‘K’ is doubled therefore rc becomes r/2

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17.

A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts.
Which of the following quantities remain constant in the wire?
(a) drift speed
(b) current density
(c) electric current
(d) electric field


The electric current will remain constant in a wire whose cross-sectional area is increased linearly from its one end to the other, is connected across a battery of V volts.

Because current is the only quantity that does not depend on the area of cross- sections of the wire.
I=dq/dt, that is the rate of flow of charge, where as drift speed, current density and electric field depending on the increasing area of the cross-section of the following relations:

Drift speed: νd=I/Ane
Current density = I/A
Electric field = J/σ



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18. Find out the wavelength of the electron orbiting in the ground state of the hydrogen atom.

Radius of ground state of hydrogen atom =0.53A°  = 0.53 x10-10m
According to de Broglie relation 2πr = nλ
For ground state n=1
2 x 3.14 x 0.53 x 10-10= 1 x λ
therefore, λ = 3.32 x10-10 m
= 3.32 A° 

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19.

A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s−1 in a uniform magnetic field of magnitude 3.0 × 10−2T. Calculate the maximum value of the current in the coil.


Here, A = 200 cm2
N=20
ω = 50 rad/s
B = 3.0× 10−2 T
for maximum current emf induced should be maximum.

So, for maximum emf sinωt in e = NAB ω sinωt should be 1.

Hence, e = NABω
e = 20 × 200/10000 × 50 × 3 × 10−2 = 0.6 V

Since resistance of the circular loop is not given, let us consider its resistance to be R.

Therefore, current I in the coil is
straight I space equals space fraction numerator 0.6 over denominator straight R end fraction space straight A

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20.

Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving a reason, the frequency range used in this mode of propagation.


There are two basic modes of communication which are used in satellite communication:

(i) point-to-point and broadcast-  In the broadcast mode, there are a large number of receivers corresponding to a single transmitter. Satellite communication is an example of the broadcast mode of communication.

(ii) Space wave mode of propagation: In the frequency range 30 MHz to 1000 Mhz, the wavelengths are in the range of 30 cm to 10 m. At these wavelengths the diffraction of waves is high and the waves lose their directional property. So frequency range of 5.925 - 6.425 GHz is used in satellite communication.

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