## Saturday, March 27, 2010

### 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 ρA and ρB, 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 hA and hB respectively, with hA > hB. The possible relation(s) between their resistivities and their masses mA and mB is(are)

(A) ρA > ρB and mA = mB

(B) ρA < ρB and mA = mB

(C) ρA > ρB and mA > mB

(D) ρA < ρB and mA < mB

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 ρA < ρB.

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 1/4th 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].