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CBSE Class 12 Physics Solved Question Paper 2017

Long Answer Type

31.

Describe the working principle of a moving coil galvanometer.

Why is it necessary to use
(i) a radial magnetic field and
(ii) a cylindrical soft iron core in a galvanometer?

Write the expression for the current sensitivity of the galvanometer.

Can a galvanometer as such be used for measuring the current? Explain.

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32.

(a) Define the term 'self-inductance' and write its S.I. unit.

(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length when a current I is set up in the outer solenoid S₂.

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33. Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.

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34.

(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turn and currents in the two coils.

(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 300 turns. Find the number of turns in the secondary to get the power output at 220 V.

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35.

Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm's law.

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36.

State the two Kirchhoff's laws. Explain briefly how these rules are justified.

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37.

The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for (i) the current draw from the cell and (ii) the power consumed in the network.

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38.
(i) A ray of light incident on face AB of an equilateral glass prism shows the minimum deviation of 30°. Calculate the speed of light through the prism.

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.

At the minimum deviation, the refracted ray inside the prism becomes parallel to its base.
Angle of minimum deviation is given as Dm = 30°
Since, the prism is equilateral, So, A = 60°
Refractive index of the prism
$straight mu space equals space fraction numerator Sin space open parentheses begin display style fraction numerator straight A space plus space straight D subscript straight m over denominator 2 end fraction end style close parentheses over denominator Sin space begin display style straight A over 2 end style end fraction space equals space fraction numerator Sin space begin display style 90 over 2 end style over denominator Sin space begin display style 60 over 2 end style end fraction space equals space fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator 1 divided by 2 end fraction space equals space square root of 2$

We know that μ = v₁/v₂
Hence the speed of light in prism would be 1/√2 times the speed of light in air i.e = 3 x10⁸ /1.414 = 2.121 x10⁸ m/s

(ii)

From Snell's law, we know that $fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space straight mu subscript 12$

For the emergent ray to graze at the face AC, the angle of refraction should be 90
So, applying snell's law at face AC, we get
$fraction numerator sin space straight i subscript AC over denominator sin space straight r subscript AC end fraction space equals space straight mu subscript 21 space rightwards double arrow space fraction numerator sin begin display style space end style begin display style begin display style straight i end style subscript AC end style over denominator sin begin display style space end style begin display style begin display style 90 end style to the power of straight o end style end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction space or space straight i subscript AC space equals space 45 to the power of straight o$

From figure, we can see that angle of refraction at face AB is 15
So applying Snell's law we get
Sin i_AB = sin r_AB x μ₁₂
or i_AB = sin^-1 (√sin 15°)

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