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IIT JEE 2009 Physics Paper2 Code 1 Solutions

6.     The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,


        (A)    the process during the path A -> B is isothermal

        (B)    heat flows out of the gas during the path B -> C -> D

        (C)    work done during the path A -> B -> C is zero

        (D)    positive work is done by the gas in the cycle ABCDA


Sol.   (B & D)

         1st law of thermodynamics

        DQ = DU + W

        For process B -> C -> D

        ΔU is negative as well as W is also negative


7.     A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then,

circular motion

Sol.   (B & C).

          Simple concept of rolling.Velocity of point A is 0


8.     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


Sol.   (A & D)

        Larger the length of air column, lesser is the intensity


9.     Two metallic rings A and B, identical in shape and size but having different resistivities rA and rB, 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


Sol.   (B & D)

        Since  - dΦ/dt = emf is same, the current induced in the ring will depend upon resistance of the ring. Greater the resistivity smaller the current.

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