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`CBSE`

Physics

CBSE Class 12

1.

How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux.

During charging, electric flux between the plates of capacitor keeps on

changing; this results in the production of a displacement current

between the plates.

Expression for the displacement current

2.

A long straight current carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will the be an induced emf in the loop? Justify

No, Magnetic flux does not change with the change of current.

3.

A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of R_{o}. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire.

When the slide is in the middle of the potentiometer, only half of its total resistance i.e. R_{0}/2 (since resistance is directly proportional to length) will be between A and point of contact (C), say R_{1}, will be given by the following expression.

The total resistance between A and B will be sum of the resistance between A & C and C&B i.e R_{1}+R_{0}/2

Current flowing through the potentiometer will be

The voltage V1 taken from the potentiometer will be the product of current I and the resistance R1

4.

Depict the orientation of the dipole in (i) stable, (ii) unstable equilibrium in a uniform electric field.

Depicting the orientation of the dipole in (i) stable equilibrium in a uniform electric field.

Net force is zero in this case as qE-qE = 0

Net torque =pEsinθ as θ = 0

(ii) Unstable equilibrium in a uniform electric field.

Net force is negative in this case as-qE-qE = -2qE

Net torque = pEsinθ asθ = 180

5.

Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give a reason for your answer.

No, Because the charge resides only on the surface of the conductor.

6.

(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2μF capacitance.

(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?

7.

(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase: current or voltage?

(ii) Without making any other change, find the value of the additional capacitor C, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

Given V = Vosin (1000 +Φ)

ω = 1000s^{-1}

L = 100 mH

C =2μF

R = 400Ω

Phase difference

X_{L} = ωL = 1000 x100 x10^{-3} = 100Ω

=

(ii) For power factor to be unity, R = Z

or X_{L} = X_{C}

For two capacitance in parallel, resultant capacitance C'=C + C1

10-5 = 0.2 x 10_{-5} + C_{1}

⇒ C_{1} = 8μF

8.

How is the speed of em-waves in vacuum determined by the electric and magnetic field?

The speed of the em-waves in a vacuum is determined by the ratio of the peak value of electric and magnetic fields.

c =E_{0}/B_{0}

9.

Drive the expression for electric field at a point on the equatorial line of an electric dipole.

The magnitudes of the electric field due to the two charges +q and -q are given by,

The directions of E+q and E-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.

Therefore, Total electric field.

E = - (E+q + E-q) cosθ p [Negative sign shows that field is opposite to p]

At large distances (r>>a),this reduces to

10.

At a place, the horizontal component of earth's magnetic field is B and angle of dip are 60^{o}. What is the value of a horizontal component of the earth's magnetic field at the equator?

The horizontal component of the electric field is given by

B_{H} = B_{e}cos(θ)

where θ is the angle of dip at the given place

Be is teh net magnetic field

At θ = 60^{o}

B_{H} = B_{e} cos(60) = B_{e} x (1/2)

B_{e} = 2BH = 2B (given BH =B)

Now horizontal component at equator

B_{H} = B_{e}cos (θ)

angle of dip at equator is 0

B_{H} = B_{e}cos(0) = Be = 2B

B_{H} = 2B