MHTCET 13 Apr 2026 Shift 1 Paper

Concept:For an ideal gas, the r.m.s. velocity is proportional to the square root of temperature, and for an adiabatic process, $T V^{\gamma-1}$ remains constant.Explanation:The r.m.s. velocity is given by $v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$.So, $v_{\mathrm{rms}} \propto \sqrt{T}$.The r.m.s. velocity is reduced to $\frac{1}{3}$ of its initial value, so $\frac{v_2}{v_1} = \frac{1}{3}$.Using $v \propto \sqrt{T}$, we get $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Thus, $\frac{1}{3} = \sqrt{\frac{T_2}{T_1}}$.Squaring both sides gives $\frac{T_2}{T_1} = \frac{1}{9}$.For an adiabatic process, $T V^{\gamma-1} = \text{constant}$.Given $\gamma = 1.5$, so $\gamma - 1 = 0.5 = \frac{1}{2}$.Therefore, $T V^{1/2} = \text{constant}$.So, $T_1 V_1^{1/2} = T_2 V_2^{1/2}$.Substitute $T_2 = \frac{T_1}{9}$: $T_1 V_1^{1/2} = \frac{T_1}{9} V_2^{1/2}$.Cancelling $T_1$, we get $V_1^{1/2} = \frac{1}{9} V_2^{1/2}$.Hence, $V_2^{1/2} = 9 V_1^{1/2}$.Squaring both sides gives $V_2 = 81 V_1$.So, the gas must be expanded 81 times.Answer:The correct option is B, 81 times.