Sunday, October 11, 2009

IIT JEE Physics Final Revision Set 8

One mole of a diatomic ideal gas (y = 1.4) is taken through a cyclic process starting from point A. The process A→B is an adiabatic compression. B→C is isobaric expansion, C→D an adiabatic expansion and D→A is isochoric. The volume ratios are VA/VB = 16 and VC/VD = 2 and the temperature at A is TA = 300 K.

1. Calculate the temperature of the gas at the point B and D

2. Find the efficiency of the cycle.

3. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is:
(A) 0.0015 (B) 0.003
(C) 0.048 (D) 0.768

4. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:
(A) 225 (B) 450
(C) 900 (D) 1800

5. A proton, a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd and rα denote respectively the radii of the trajectories of these particles, then:
(A) rα = rp is l.t. rd (B) rα > rd > rp
(C) rα = rd is l.t. rp (D) rp = rd = rα

Source 1997 JEE

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