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(a) – 5 J (b) – 10 J (c) – 15 J (d) – 20 J Answer-a (118)
(a) 34 J (b) 70 J (c) 84 J (d) 134 J Answer-d (202)
(a) BA, AD, DC (b) DC, CB, BA (c) AB, BC, CD (d) CD, DA, AB Explanation-(d) Process CD is isochoric as volume is constant, Process DA is isothermal as temperature constant and Process AB is isobaric as pressure is …
(a) C and D respectively (b) D and C respectively (c) A and B respectively (d) B and A respectively Answer-c (105)
(a) He and O2 (b) O2 and He (c) He and Ar (d) O2 and N2 Answer-b (138)
PA=3×104 Pa and PB=8×104 Pa, VA=2×10-3 m3 and VD=5×10-3 m3 . In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy …
β=-(dV/dP)/V with P for an ideal gas at constant temperature Answer-a (108)
Q1,Q2 and Q3 denote the heat absorbed by the gas along the three paths, then (a) Q1 < Q2 < Q3 (b) Q1 < Q2 = Q3 (c) Q1 =Q2 > Q3 (d) Q1 > Q2 > Q3 Answer-a (98)
(a) ac (b) cg (c) af (d) cd Explanation-(c) Area enclosed between a and f is maximum. So work done in closed cycles follows a and f is maximum. (114)
(a) 2PV (b) PV (c) 1/2PV (d) Zero Answer-a (45)
(a) 20 J (b) – 20 J (c) 400 J (d) – 374 J Ans. b (62)
ΔU1 and ΔU2 are the changes in internal energies in the processes I and II respectively, then (a) ΔUII > ΔUI (b) ΔU1 < ΔU2 (c) ΔU1 = ΔU2 (d) Relation between ΔUI and ΔUII can not be determined Ans. (c) …
(a) 30 K (b) 18 K (c) 50 K …
(Cp/Cv=γ) is at pressure P1 and temperature T1 in left part and gas at pressure P2 and temperature T2 in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure …
(a) 4 R (b) 2.5 R (c) 3 R (d) 4R/3 Answer-c (9)
PV2= constant. During this process, the gas is (a) Heated (b) Cooled (c) Neither heated nor cooled (d) First heated and then cooled Answer-b (7)
(a) – 4.5 cal/K (b) + 4.5 cal/K (c) +5.4 cal/K (d) – 5.4 cal/K Answer-b (18)
(a) Valid (b) Invalid (c) Depends on engine design (d) Depends of the load Answer-b (2)
(a) 80°C, 37°C (b) 95°C, 28°C (c) 90°C, 37°C (d) 99°C, 37°C Answer-d (4)
(a) 0 kJ (b) 10 kJ (c) 20 kJ (d) 30 kJ Answer-d (8)
(a) Greater than the final pressure of B (b) Equal to the final pressure of B (c) Less than the final pressure of B (d) Twice the final pressure of B Answer-c (9)
(a) 2 moles of helium occupying 1m3 at 300 K (b) 56 kg of nitrogen at 107N/m2 and 300 K (c) 8 grams of oxygen at 8 atm and 300 K (d) 6×1026 molecules of argon occupying 40m3 at …
(a) 40 % (b) 30 % (c) 60 % (d) 20 % Answer-a (8)
T1=800K and rejects to sink at T2K . The second engine B receives heat rejected by the first engine and rejects to another sink at T3=300K. If the work outputs of two engines are equal, then the value of T2 …
(a) PV3/2 = constant (b) PV5/2 = constant (c) PV7/3 = constant (d) PV4/3 = constant Answer-a (5)
(a) 5% (b) 6% (c) 7% (d) 8% Answer-c (2)
(a) (2/5)R (b) (5/2)R (c) (10/3)R (d) (6/7)R Ans. c (9)
temperature is T0, atmospheric pressure is also P0. Now the temperature of the gas is increased to 2T0, the tension in the wire will be (a) 2P0A …
volume V1 to V2 for a gas which obeys Vander Waal’s equation (V – βn)[ P + (αn2/v)= nRT (a) nRT loge( V2 – nβ/V1– nβ) + α n2(V1 -V2)/V1V2 (b) nRT log10 ( V2 – αβ/V1– αβ) + α n2(V1 -V2)/V1V2 …
(a) 4 RT (b) 15 RT (c) 9 RT (d) 11 RT Ans. d (6)
(a) PV3/2 = constant (b) PV5/2 = constant (c) PV7/3 = constant (d) PV4/3= constant Answer-a (7)
(a) 5% (b) 6% (c) 7% (d) 8% Answer-c (4)
(a) (2/5)R (b) (5/2)R (c) (10/3)R (d) …
P0 and temperature is T0, atmospheric pressure is also P0. Now the temperature of the gas is increased to 2T0, the tension in the wire will be (a) 2P0 A …