JEE Main 2019 — Kinetic Theory of Gases Question with Solution

Heat loss $=$ Heat gain $=\mathrm{mS}\Delta \theta$ $\mathrm{So},{\mathrm{m}}_{\mathrm{A}}{\mathrm{S}}_{\mathrm{A}}\Delta {\theta }_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{S}}_{\mathrm{B}}\Delta {\theta }_{\mathrm{B}}$ $\begin{array}{l} \Rightarrow 100 \times \mathrm{S}_{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}_{\mathrm{B}} \times(90-75) \\ 2 \mathrm{~S}_{\mathrm{A}}=1.5 \mathrm{~S}_{\mathrm{B}} \Rightarrow \mathrm{S}_{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}} \end{array}$ Now, $100\times {\mathrm{S}}_{\mathrm{A}}\times (100-\theta )=50\times {\mathrm{S}}_{\mathrm{B}}\times (\theta -50)$ $2\times \left(\frac{3}{4}\right)\times (100-\theta )=(\theta -50)$ $\begin{array}{l} 300-3 \theta=2 \theta-100 \\ 400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C} \end{array}$