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# CBSE Physics 2013 Exam Questions

1.

A capacitor has been charged by a dc source. What are the magnitudes of conduction and displacement currents, when it is fully charged?

When capacitor is charged by a dc source,

Conduction current is equal to the displacement current.

i.e.,
Because,  is maximum when, capacitor is fully charged.

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2.

The emf of a cell is always greater than its terminal voltage. Why? Give reason.

V= E – I r;

The emf of a cell is greater than terminal voltage because there is some potential drop across the cell due to its small internal resistance.

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3.

State Lenz’s Law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.

The polarity of the induced emf at the open ends of a closed loop is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

Emf will be induced at the ends. Magnetic flux changes in a metallic rod falling freely under gravity and hence, emf is induced.

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4.

What ate permanent magnets? Give one example.

Substances that can retain their magnetism for a long duration of time at room temperature are called permanent magnets. They have high retentivity and coercivity.

Eg: Steel, cobalt

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5.

An ammeter of resistance 0.80 can measure current upto 1.0A.

(i) What must be the value of shunt resistance to enable the ammeter to measure current upto 5.0A?

(ii) What is the combined resistance of the ammeter and the shunt?

Given, Resistance, R = 0.80

Maximum current, Imax = 1 A

So, Voltage across ammeter, V = IR = 1.0 0.80 = 0.8 V

i) If a shunt is connected in parallel,

Current I flowing through S is (I – I1)

For parallel combination of resistors,

I1. RA = ((I – I1) S

This is the required value of shunt resistance.

ii) Combined resistance of the ammeter and the shunt,

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6.

What is the geometrical shape of equipotential surfaces due to a single isolated charge?

The equipotential surface for an isolated charge is concentric circles. The distance between the shell decreases with increase in electric field.

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7.

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38  as shown in the figure. Find the value of the current in circuit.

Since the cells are in opposition,

E net = E1 – E2 = (200-10) = 190 V

So, current in the circuit is given by, I =

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8.

(a) An em wave is traveling in a medium with a velocity . Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.

(b) How are the magnitudes of the electric and magnetic fields related to the velocity of the em wave?

represents the direction of propagation of wave. The figure below shows us the propagation of the em wave.

a) The direction of oscillating field is
b) Speed of EM wave is given by |c| =  ; Eo is the magnitude of electric field and is the magnitude of magnetic field.
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9.

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

Given, a convex lens placed coaxially in contact with a concave lens.

Power of convex lens, P1
Power of concave lens, P2

So, Power of combination, P = P1 + P

=

Since, power has negative magnitude; system of lenses is diverging in nature.

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10.

A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 C . When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 C .

Calculate:

(i) The potential V and the unknown capacitance C.

(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120V?

OR

A hollow cylindrical box of length 1m and area of cross-section 25cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by, where E is in NC-1 and x is in metres. Find:

(i) Net flux through the cylinder.

(ii) Charge enclosed by the cylinder.

The charge on an unknown capacitor is given by,

Q = CV

CV = 360 C                                  ... (1)

On reducing the potential by 120 V, charge on the capacitor is reduced which is given by,

Q’ = C(V-120)

C(V-120) = 120 C                          … (2)

On solving equation (1) and (2), we have

Unknown capacitance from equation (1),

Q = CV

ii) Charge on the capacitor, if voltage is increased by 120V

Q = C (V+120)

= 2 (180+120)

= 600 C

OR

i) Electric flux through a surface,
Flux through the left surface,

Since x = 1m,

ii) Using Gauss Theorem, we can calculate the charge inside the cylinder.

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