Physics

CBSE Class 12

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1.

a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.

b) A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of paper, as shown. Trace their paths in the field and justify your answer.

a)

Force acting on a charged particle q, which is moving with velocity v in a magnetic field B, is given by,

The right-hand rule gives the direction of this force. The direction of the force is perpendicular to the plane containing velocity v and magnetic field B.

b)

A charged particle experiences a force when it enters the magnetic field. Due to the presence of magnetic field, the charged particle will move in a circular path. This is because the force is perpendicular to the velocity of the charged particle.

Radius of the circular path in which the charged particle is moving is given by,

Since the neutron has no charge, it will move along a straight line.

The electron will follow a circular path which has a radius smaller than that of the alpha particle. This is because the mass to charge ratio of the alpha particle is more than that of the electron.

Therefore, the electron will move in the clockwise direction and the electron will move in the anticlockwise direction as per the Right Hand Rule.

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2.
Why are microwaves considered suitable for radar systems used in aircraft navigation?

Microwaves are considered suitable for radar systems used in aircraft navigation because they have a short wavelength range (10

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3.

i) Define mutual Inductance.

ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes 20 A in 0.5 s, what is the change of flux linkage with the other coil?

i) Mutual Induction: It is the phenomenon in which a change of current in one coil induces an emf in another coil places near it. The coil in which the current changes is called the primary coil and the coil in which the emf is induced is called the secondary coil.

ii)

EMF induced is, e = - M .

So, the flux linked with the other coil is given by,

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4.
Two cells of emfs 1.5 V and 2.0 V, having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

Given,

EMF, E_{1} = 1.5 V ; E_{2} = 2 V

Internal resistance, r = 0.2

Effective emf of two cells connected in parallel is,

E_{eff} =

This implies,

The effective resistance can be calculated as:

R_{eff} =

That is,

R_{eff} =

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5.
Write the underlying principle of a moving coil galvanometer.

When a current-carrying coil is placed in a magnetic field, it experiences torque. This is the underlying principle of a moving coil galvanometer.

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6.
Two parallel-plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates, while Y contains a dielectric medium of

i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4

ii) Calculate the potential difference between the plates x and Y

iii) Estimate the ratio of electrostatic energies stored in x and Y

i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4

ii) Calculate the potential difference between the plates x and Y

iii) Estimate the ratio of electrostatic energies stored in x and Y

i)

Capacitance of a parallel plate capacitor is given by,

ii)

Potential difference between plates X and Y can be calculated as,

Q = CV

iii)

The ratio of electrostatic energies can be calculated as:

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7.

A point charge +Q is placed at point O as shown in the figure. Is the potential

difference V_{A} – V_{B} is positive, negative or zero?

Electric potential at a distance r from the point charge +Q is given by,

Potential at point A is,

Similarly, potential at point B is given by,

Since r_{A} < r_{B}

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8.
A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence, show that for points at large distance from the ring, it behaves like a point charge.

Consider a ring of radius 'a' which carries uniformly distributed positive total charge Q.

To find: electric field due to a ring at a point P lying at a distance x from its centre along the central axis perpendicular to the plane of the ring.

As the charge is distributed uniformly over the ring, the charge density over the ring is,

The perpendicular component of electric field due to charge on the ring along the x-axis cancels each other out.

As there is same charge on both sides of the ring, the magnitude of the electric field at P due to the segment of charge dQ is given by,

dE = k_{e}

E_{x} =

1. At the centre (X = 0) , electric field is zero.

2. When x>> a, a can be neglected in the denominator.

Therefore,

E

1. At the centre (X = 0) , electric field is zero.

2. When x>> a, a can be neglected in the denominator.

Therefore,

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9.
Define 'quality factor' of resonance in a series LCR circuit. What is its SI unit?

Quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation.

Quality factor is expressed as,

Q-factor is a dimensionless quantity.

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10.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

According to Gauss's law, flux through a closed surface is given by,

Here, q is the charge enclosed by the gaussian surface.

That is, on increasing the radius of the gaussian surface, charge q remains unchanged.

So, flux through the gaussian surface will not be affected when its radius is increased.

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Textbook Solutions | Additional Questions

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