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ISC Class 12 Physics Solved Question Paper 2009

Short Answer Type

21.

(i) What is meant by:

(1) Drift velocity and

(2) Relaxation time?

(ii) A metallic plug AB is carrying a current I (see Figure 4 below). State how the drift velocity of free electrons varies, if at all, from end A to end B.

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22. Figure below shows a uniform manganin wire XY of length 100 cm and resistance 9 Ω, connected to an accumulator D of emf 4 V and internal resistance 1 Ω through a variable resistance R. E is sa cell of emf 1.8 V connected to the wire XY via a jockey J and a central zero galvanometer G. It is found that the galvanometer shows no deflection when XJ = 80 cm. Find the value of R.

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23. Obtain an expression for magnetic flux density ‘B’ at the center of a circular coil of radius R and having N turns, when a current I is flowing through it.

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24.

(i) State any two differences between a moving coil galvanometer and a tangent galvanometer.

(ii) What is the use of a Cyclotron?

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25. What is meant by the time constant of an LR circuit? When the current flowing through a coil P decreases from 5A to 0 in 0.2 seconds, an emf of 60 V is induced across the terminals of an adjacent coil Q. Calculate the coefficient of mutual inductance of the two coils P and Q.

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26.

When an alternating emf e = 300 Sin (100 πt + π/6) Volt is applied to a circuit, the current I through it is I = 5.0 Sin (100 πt - π/6) ampere. Find the :

(i) Phase difference between the emf and the current.

(ii) Average power consumed by the circuit.

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27. Obtain an expression for resonant frequency ‘f₀’of a series LCR circuit.

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28.

In which part of the electromagnetic spectrum, do the following radiations lie:

(i) Having wavelength of 20 nm

(ii) Having frequency of 10 MHz

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29. A thin convex lens which is made of glass (refractive index 1.5) has a focal length of 20 cm. It is now completely immersed in a transparent liquid having refractive index 1.75. Find the new focal length of the lens.

Given, Focal length of convex lens = 20 cm

Refractive Index of convex lens =1.5

Refractive index of liquid = 1.75

$straight I subscript μg space equals space straight mu subscript straight g over straight mu subscript straight c space equals space fraction numerator 1.5 over denominator 1.75 end fraction space equals space 6 over 7$

For air,

$1 over straight f subscript straight a space equals space left parenthesis straight mu presubscript straight a subscript straight g space minus space 1 right parenthesis space open parentheses 1 over straight R subscript 1 space minus space 1 over straight R subscript 2 close parentheses 1 over straight f subscript straight g space equals space open parentheses straight mu presuperscript straight a subscript straight g space minus space 1 close parentheses open parentheses 1 over straight R subscript 3 space minus space 1 over straight R subscript 2 close parentheses straight f subscript straight g over straight f subscript straight a space equals space fraction numerator straight mu presuperscript straight a subscript straight g space minus space 1 over denominator straight mu presubscript straight a subscript straight g space minus space 1 end fraction space equals space fraction numerator 1.5 space minus space 1 over denominator 9 divided by 8 space minus space 1 end fraction space equals space 4$

Therefore,

f_e = 4 x f_a = 4 x 20 = 80 cm

New focal length = f_{new =}_{$1 over straight f subscript new space equals space 1 over f subscript g space plus space 1 over f subscript e space equals space 1 over 20 space plus space 1 over 80$}f_new = 16 cm