Physics Delhi (Set 3)

General Instructions:

Read the following instructions very carefully and strictly follow them:

1. This question paper comprises four sections – A, B, C and D.
2. There are 37 questions in the question paper. All questions are compulsory.
3. Section A: Q. no. 1 to 20 are very short-answer type questions carrying 1 mark each.
4. Section B : Q. no. 21 to 27 are short-answer type questions carrying 2 marks each.
5. Section C : Q. no. 28 to 34 are long-answer type questions carrying 3 marks each.
6. Section D : Q. no. 35 to 37 are also long answer type questions carrying 5 marks each.

1. There is no overall choice in the question paper. However, an internal choice has

been provided in two questions of one mark, two questions of two marks, one question of three marks and all the three questions of five marks. You have to attempt only one of the choices in such questions.

1. However, separate instructions are given with each section and question, wherever necessary.
2. Use of calculators and log tables is not permitted.
3. You may use the following values of physical constants wherever necessary.

<math>c = 3 \times 10^8 \,\text{m/s}</math>

<math>h = 6.63 \times 10^{-34} Js</math>

<math>e = 1.6 \times 10^{-19} C</math>

<math>\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}</math>

<math>\epsilon_0 = 8.854 \times 10^{-12} \,\mathrm{C}^2 \,\mathrm{N}^{-1} \,\mathrm{m}^{-2}</math>

<math>4\pi\varepsilon_0 = 9 \times 10^9 \,\mathrm{N} \,\mathrm{m}^2 \,\mathrm{C}^{-2}</math>

Mass of electron (m_e) = <math>9.1 \times 10^{-31}</math> kg

Mass of neutron = <math>1.675 \times 10^{-27}</math> kg

Mass of proton = <math>1.673 \times 10^{-27}</math> kg

Avogadro's number = <math>6.023 \times 10^{23}</math> per gram mole

Boltzmann constant = <math>1.38 \times 10^{-23} \text{ JK}^{-1}</math>

Question 1

A biconcave lens of power P vertically splits into two identical plano concave parts. The power of each part will be

1. 2P
2. P/2
3. P
4. P/√2

SOLUTION:

Let the focal of the biconcave lens be f and that of the plano-concave lens be <math>f'</math>.

Let the radius of curvature for the biconcave lens be R. Applying Lens maker's formula

<math>\frac{1}{f} = (\mu - 1) \left( \frac{1}{(-R)} - \frac{1}{R} \right) = -\frac{2(\mu - 1)}{R}</math>

For the plano-concave lens, the focal length can be calculated as:

<math>\frac{1}{f'} = (\mu - 1) \left( \frac{1}{(-R)} \right) = -\frac{(\mu - 1)}{R}</math>

Thus, <math>f' = 2f</math>

Thus, power of the plano-concave lens will be P'=P/2 Hence, the correct answer is option (b)

Question 2

The power factor of a series LCR circuit at resonance will be

1. 1
2. 0
3. 1/2
4. √12

SOLUTION: At resonance condition, the alternating current and voltage are in same phase, thus the phase difference between them <math>\Phi = 0</math>, and the power factor, <math>\cos \Phi = 1</math>. Hence the correct answer is option (a).

Question 3

If photons of frequency v are incident on the surfaces of metals. A & B of threshold frequencies v/2 and v/3 respectively, the ratio of the maximum kinetic energy of electrons emitted from A to that from B is

1. 2:3
2. 3:4
3. 1:3
4. √3 : √2

SOLUTION: According to Einstein's photoelectric equation,