IIT-JEE 2009- Multiple Correct Choice Type Questions on Electromagnetic Induction, Resonance Column & Kepler’s Laws

Today we will discuss three Multiple Correct Choice Type questions which appeared in the IIT-JEE 2009 question paper. Even though these questions are named ‘multiple correct choice’ type the number of correct choices can be one or more. These questions are simple, in line with the current trend, as you can see:

(1) Two metallic rings A and B, identical in shape and size but having different resistivities ρ_A and ρ_B, are kept on top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights h_A and h_B respectively, with h_A > h_B. The possible relation(s) between their resistivities and their masses m_A and m_B is(are)

(A) ρ_A > ρ_B and m_A = m_B

(B) ρ_A < ρ_B and m_A = m_B

(C) ρ_A > ρ_B and m_A > m_B

(D) ρ_A < ρ_B and m_A < m_B

The voltages induced in the rings are equal. Since ring A jumps to greater height compared to ring B, the current in ring A must be greater. This means that the resistivity of ring A is less than that of ring B. So ρ_A < ρ_B.

Ring A can jump to a greater height if the mass of ring A is equal to or less than that of B. So the correct options are (B) and (D).

(2) A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then

(A) the intensity of the sound heard at the first resonance was more than that at the second resonance

(B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube

(C) the amplitude of vibration of the ends of the prongs is typically around 1 cm

(D) the length of the air-column at the first resonance was somewhat shorter than 1/4^th of the wavelength of the sound in air.

Option (A) is correct. (The second resonance will give sound of smaller intensity since the air column is longer and hence there is greater chance for the attenuation of the sound before and after the reflection at the water surface).

Option (B) is incorrect since the prongs are to be kept in a vertical plane to transfer energy efficiently into the resonance column.

Option (C) also is incorrect since the amplitude of vibration of the ends of the prongs is typically of the order of a millimetre only.

Option (D) is correct since the anti-node is obtained slightly above the end of the resonance tube and λ/4 = l + e where λ is the wave length of sound, l is the resonating length of the tube and e is the end correction.

(3) Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s)

(A) the angular momentum of the charge –q is constant

(B) the linear momentum of the charge –q is constant

(C) the angular velocity of the charge –q is constant

(D) the linear speed of the charge –q is constant.

The situation here is similar to that of a hydrogen atom (or, the motion of a planet around the sun) which is central field motion under an inverse square law force. The only correct option is (A) which high lights the constancy of angular momentum (Kepler’s law).

[Note that the torque is zero and hence the angular momentum is constant].

Questions (MCQ) on Oscillations of a Spring-mass System

The following questions involving a spring-mass system are meant for checking whether you have a thorough understanding of the simple harmonic oscillations in which the restoring force is supplied by a spring:

(1) The frequency of vertical oscillations of a mass suspended at the end of a light spring is n. If the system is taken to a location where the acceleration due to gravity is reduced by 0.1%, the frequency of oscillation will be

(a) 1.01 n

(b) 0.99 n

(c) 1.001 n

(d) 0.999 n

(e) n

A spring-mass system (unlike the simple pendulum) does not require a gravitational force for oscillations since the restoring force required for oscillations is supplied by the elastic forces in the spring.

[Note that the period (T) of oscillations is given by T = 2π√(m/k) where m is the mass attached to the spring and k is the spring constant].

Therefore the change in ‘g’ does not affect the frequency and the correct option is (e).

(2) One end of a light spring is fixed to the ceiling and a mass M is suspended at the other end. When an additional mass m is attached to the mass M, the additional extension in the spring is e. The period of vertical oscillation of the spring-mass system now is

(a) 2π√[(M+m)e/mg]

(b) 2π√(me/mg)

(c) 2π√[(M+m) /mge]

(d) 2π√[me/(M+m)g]

(e) 2π√(M/mge)

The period (T) of oscillations is given by T = 2π√[(M+m)/k] where k is the spring constant.

Since an additional weight mg attached to the spring produces an additional extension e, the spring constant k = mg/e.

Therefore, period of oscillations T = 2π√[(M+m)/(mg/e)] = 2π√[(M+m)e/mg], as given in option (a).

(3) The period of vertical oscillations of a mass M suspended using a light spring of spring constant k is T. The same spring is cut into three equal parts and they are used in parallel to suspend the mass M as shown in the adjoining figure. What is the new period of oscillations?

(a) T

(b) 3T

(c) 9T

(d) T/9

(e) T/3

The original period of oscillation T is given by

T = 2π√(M/k)

When the spring is cut into three equal parts, each piece has spring constant 3k.

[Since the length of each piece is reduced by a factor three, the extension for a given applied force will be reduced by a factor three so that the spring constant (which is the ratio of force to extension) will become three times].

Since the three pieces are connected in parallel the effective spring constant of the combination is 3k+3k+3k = 9k. The new period of oscillations T₁ is given by

T₁ = 2π√(M/9k) = T/3 since T = 2π√(M/k)

You will find some multiple choice questions with solution in this section here.