(i) P1 , V to 2P1 , V (ii) P1, V to P, 2V. Then work done in the two cases is (a) Zero, Zero (b) Zero, PV1 (c) PV1 , Zero (d) PV1, P1 V1 Ans. b (11)
(a) -50 joules (b) 20 joules (c) 30 joules (d) 50 joules Ans. d (5)
(a) Momentum (b) Energy (c) Mass (d) Temperature Ans. b (5)
(a) Decreases (b) Increases (c) Remains constant (d) Depends on the molecular motion Ans. c (16)
(a) – 50 J (b) 20 J (c) 30 J (d) 50 J Ans. b (6)
(a) Q + W (b) Q – W (c) Q (d) (Q – W)/2 Ans. b (2)
area 0.05 m2. Increase in its potential energy will be (T = 0.2 N/m) (a) 5 x 10-2J (b) 2 x 10-2j (c) 3 …
(a) 24πR2S (b) 48 πR2S (c) 12πR2S (d) 36πR2S Ans. a (5)
(a) dQ = dU + PdV (b) dQ = dU x Pdv (c) dQ = (dU + dv)P (d) dQ = PdU + dV Ans. a (10)
(a) r (b) 0 (c) Infinity (d) 1/2r Ans. c (13)
(a) 3 J (b) 6.5 J (c) 1.5 J (d) 4 J Ans. a (7)
(a) 192π x 10-4 J (b) 280π x 10-4 J (c) 200π x 10-3 J (d) None of these Ans. a (12)
(a) 1:21/3 (b) 21/3:1 (c) 2 : 1 (d) 1 : 2 Ans. b (10)
the increase in surface energy. (Surface tension of mercury is 0.465 J/m2 ) (a) 23.4 μJ (b) 18.5 μJ (c) 26.8 μJ …
(a) 0.56×104 J (b) 1.3×102 J (c) 2.7×102 J …
6.21×10-21 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (a) 12.42×1021 J,968 m/s (b) 8.78×1021 J,684 m/s (c) 6.21×1021 J,968 m/s …
how much work will have to be done (Surface tension of water =7.2 x10-2 N/m) (a) 7.2 x 10-6 Joule (b) 1.44 x 10-5 Joule (c) 2.88 x 10-5 Joule (d) 5.76 x 10-5 Joule …
(a) Released (b) Absorbed (c) Both (a) and (b) (d) None of these Ans. a (12)
108 drops of equal size. The energy expended in joules is (surface tension of Mercury is 460 x 10-3N/m (a) 0.057 …
(a) 2 x 10-2Nm-1 (b) 2 x 10-4Nm-1 (c) 2 x 10-6Nm-1 (d) 2 x 10-8Nm-1 …
(a) 1 (b) 2 (c) 4 (d) 6 Ans. c (10)
a ring of area is (Surface tension of liquid = 5Nm-1) (a) 0.75 J (b) 1.5 J (c) 2.25 J (d) 3.0 J Ans. b (10)
(a) 1 : 10 (b) 1 : 15 (c) 1 : 20 (d) 1 : 25 Ans. c (5)
(a) Energy is released (b) Energy is absorbed (c) The surface area of the bigger drop is greater than the sum of the surface areas of both the drops (d) The surface area of the bigger drop is smaller than …
(a) Less than 1 eV (b) A few keV (c) 50-60 eV (d) 13.6 eV Answer-a (11)
(a) 137°C (b) 127°C (c) 100°C (d) 105°C Answer-a (8)
(a) Becomes zero (b) Becomes maximum (c) Becomes minimum (d) Remains constant Answer-a (8)
(a) (1/2)PV (b) (3/2)PV (c) (5/2)PV (d) 3PV Answer-b (7)
(a) 16 : 1 (b) 1 : 8 (c) 8 : 1 (d) 1 : 1 Answer-d (20)
(a) Double (b) Half (c) One fourth (d) Four times Answer-d (4)
H2 and He gases. Which of the following statements are corrects (a) The average translational kinetic energies of H2 molecules and He atoms are same (b) The average energies of H2 molecules and He atoms are same (c) H2 molecules have greater …
(a) Total number of molecules (b) Average kinetic energy (c) Root mean square velocity (d) Mean free path Answer-b (11)
(a) Kinetic energy (b) Potential energy (c) Vibrational energy (d) Density Answer-a (4)
(a) (nkT)/N (b) (nkT)/(2N) (c) (nkT)/2 …
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 …
6.21×10-21 J . Its average kinetic energy at 227°C will be (a) 52.2×10-21 J (b) 5.22×10-21 J (c) 10.35×10-21 J …
E1 . If the temperature is increased to 327°C, then kinetic energy would be (a) 2E1 (b) (1/2)E1 …
(a) (3/2)RT (b) (2/3)RT (c) (1/2)RT …
(a) All are correct (b) I and IV are correct (c) IV is correct (d) None of these Answer-d (7)
(a) 54°C (b) 300 K (c) 327°C (d) 108°C Answer-c (6)