MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Cylinder B is heated at constant volume, while cylinder A is heated at constant pressure. Since the same heat is supplied, the heat equations are equated.Explanation:For cylinder B, the piston is fixed, so volume remains constant.Heat supplied is:$$Q = n C_V \Delta T_B$$Given $\Delta T_B = 63\ \mathrm{K}$, we get:$$Q = n C_V \times 63$$For cylinder A, the piston is free to move, so pressure remains constant.Heat supplied is:$$Q = n C_P \Delta T_A$$Since the same amount of heat is given to both cylinders:$$n C_P \Delta T_A = n C_V \times 63$$Hence:$$\Delta T_A = \frac{C_V}{C_P} \times 63$$For an ideal diatomic gas:$$C_V = \frac{5}{2}R, \quad C_P = \frac{7}{2}R$$Therefore:$$\Delta T_A = \frac{\frac{5}{2}R}{\frac{7}{2}R} \times 63$$$$\Delta T_A = \frac{5}{7} \times 63 = 45\ \mathrm{K}$$Answer:The rise in temperature of the gas in cylinder A is $45\ \mathrm{K}$.Correct option: C.