# CBSE [All India]-Set-2-2008

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• Q1

State the reason, why GaAs is most commonly used in making of a solar cell.

Marks:1

Because GaAs is a good absorber of infrared radiations which are responsible for heating effect. This is a thin p-n junction device so that incident light reaches easily.

• Q2

Two lines, A and B, in the plot given below show the variation of de Broglie wavelength,  versus 1/V where, V is the accelerating potential difference, for two particles carrying the same charge.

Which one of the two represents a particle of heavier mass?

Marks:1

Line B represents the particle of heavier mass because of its smaller slope.

• Q3

If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?

Marks:1

The electric flux depends only upon the amount of charge enclosed by the surface. Therefore, even on doubling the radius of the Gaussian surface, the value of electric flux remains unchanged.

• Q4

Why does the bluish colour predominate in a clear sky?

Marks:1

Because molecules in air scatter blue light of short wavelength from the sun which is a strong receptor in the retina of human eye, giving us the colour vision.

• Q5

Why should the spring / suspension wire in a moving coil galvanometer have low torsional constant?

Marks:1

As the instrument is very sensitive; its restoring torque per unit twist is small. Also it has a great tensile strength and is a rusting resistance.

• Q6

The instantaneous current and voltage of an a.c. circuit are given by

i = 10sin314 t A    and   v = 50sin 314 t V.

What is the power dissipation in the circuit?

Marks:1

Irms value of current= 10/

Vrms value of voltage = 50/

Phase difference = 00

Power dissipation P = Irms Vrms Cos

= 10/50/1 ( Cos00 = 1)

= 250W

• Q7

How does the angle of minimum deviation of a glass prism of refractive index 1.5 change, if it is immersed in a liquid of refractive index 1.3?(angle of prism = 600)

Marks:1

Here ag = 1.5,  aw = 1.3

Angle of prism= 600
wg

wg

Sin = wg  sin(A/2)

= 1.15  Sin300

= 1.15  0.5

= 0.575

= Sin-1 (0.575)

= 35.10

(A + m) = 70.20

m = 70.20 – 600 = 10.20

• Q8

State two characteristic properties of nuclear force.

Marks:1

Two characteristic properties of nuclear force:

(i) Strongest attractive forces in nature.

(ii) They have charge independent character.

• Q9

Two metallic wires of the same material have the same length but cross-sectional area is in the ratio 1:2. They are connected (i) in series and (ii) in parallel. Compare the drift velocities of electrons in the two wires in both the cases (i) and (ii).

Marks:2

It is given,  A1= A and A2 = 2A

Case (i) when the wires are connected in series

Case (ii) when the wires are connected in parallel

But in parallel the current is divided in both wire.

If the current flow in the A wire is I the in the wire B is 2I,

• Q10

State one feature by the phenomenon of interference can be distinguished from that of diffraction.
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2nd order minimum from the centre of the screen is 9.5 mm. calculate the width of the slit.

Marks:2

Following are the important points of difference between interference and diffraction of light.
1.Interference is due to superposition of two distinct waves coming from two coherent sources. Diffraction is produced as result of superposition of the secondary wavelets coming from different parts of the same wave front.
2.In interference pattern, all the bright fringes (or bands) are of same intensity. In diffraction pattern, all the bright bands are not of the same intensity.

Numerical: Here  = 600nm = 600  10-9 m

X = 9.5mm = 9.5  10-3m , D = 0.8 m,

d = a = ?

For 2nd minima

X = 2D/d

d

= 101.05  10-6 m

= 1.01  10-4 m.