# CBSE [All India]-Set-3-2004

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• Q1

Marks:1

Shortwave band is used for long distance radio broadcast since it can be easily reflected back to earth by ionosphere layer of our atmosphere.

• Q2

Peak value of emf of an a.c. source is E0. What is its r.m.s. value?

Marks:1

The rms value of e.m.f is given by the relation,

E r.m.s = Eo

• Q3

Draw the voltage current characteristic of a zener diode.

Marks:1

The voltage current characteristics of a zener diode is given below,

• Q4

Why is ground wave transmission of signals restricted to a frequency of 1500 kHz?

Marks:1

Ground wave transmission is restricted to a frequency of 1500 kHz since above this frequency the attenuation of these signals increases rapidly. This occurs due to its absorption by atmosphere.

• Q5

Draw the voltage-current characteristic of a p-n junction diode in forward bias.

Marks:1

The voltage-current characteristic of a p-n junction diode under forward bias condition is given below,

• Q6

A T.V. tower has a height of 400 m at a given place. Calculate as coverage range, if the radius of the earth is 6400 km.

Marks:2

The coverage range is given by the relation,

• Q7

Explain how does the resistivity of a conductor depend upon (i) number density ‘n’ of free electrons, and (ii) relaxation time ’.

Marks:2

The resistivity of the conductor is given by the relation,
= m / (n e2)

(i) From the above equation it follows that resistitvity of a given conductor is inversely to the number density ‘n’ of free electrons.

(ii) The resistivity of the conductor increases with decrease in relaxation time and vice-versa. This is because it is inversely proportional to relaxation time.

• Q8

A circular coil of 200 turns, radius 5cm carries a current of 2.5 A. It is suspended vertically in a uniform horizontal magnetic field of 0.25 T, with the plane of the coil making an angle of 600 with the field lines.
Calculate the magnitude of the torque that must be applied on it to prevent it from turning.

Marks:2

Given that, N=200, r=5cm=0.05m, I=2.5 A, B=0.25 T,
= 90 - 60 = 30

= NIBA sin
= 200 x 2.5 x 25 x (3.14 x 0.05 x 0.05) x sin 30= 0.49 Nm

• Q9

A bar magnet M is dropped so that it falls vertically through the coil C. The graph obtained for voltage produced across the coil vs time is shown in figure (b).
(i) Explain the shape of the graph.
(ii) Why is the negative peak longer than the positive peak?

Marks:2

(i) As the magnet falls it increases the flux linked with the coil C as a result an induced e.m.f is produced. When the magnet is at the central position w.r.t coil the rate of change of flux is zero and hence the e.m.f induced for this position is zero. Again as the magnet comes out of the coil the flux linked with the coil decreases because of which the the e.m.f again starts to induce in the coil but in the opposite direction. As the magnet moves far away from the coil the flux linked with the coil becomes zero and e.m.f induced in the coil also becomes zero.
(ii) The magnet moves out of the coil more rapidly than it enters the coil. This is the reason why negative peak is greater than positive peak.

• Q10

Draw a ray diagram of an astronomical telescope in the normal adjustment position. What will be the length of the telescope for this position?

Marks:2