AP EAPCET 19-Aug-2021 Shift 1 Paper

Given, initial temperature $T_{1}=30^{\circ}\mathrm{C}$ $=303\ \mathrm{K}$Change in radius, $\Delta r=0.08\ \mathrm{mm}$Internal diameter of cylinder $d_{c}=15\ \mathrm{cm}$ $\therefore$ Internal radius of cylinder, $r_{c}=7.5 \times 10^{-2}\ \mathrm{m}$Radius of piston, $r_{p}=(7.5-0.008) \times 10^{-2}\ \mathrm{m}$ $=7.492 \times 10^{-2}\ \mathrm{m}$Linear expansion of cylinder $\alpha_{C}$ and piston $\alpha_{p}$ are $1.2 \times 10^{-5}/^{\circ}\mathrm{C}$ and $1.6 \times 10^{-5}/^{\circ}\mathrm{C}$ .Let, $\Delta r_{C}$ and $\Delta r_{p}$ is increase in radius of cylinder and piston respectively, for fully filled piston in cylinder.By using equation of linear expansion, $r=r_{0}(1+\alpha \Delta T)$ $\therefore \;\; \Delta r_{C}=r_{C} \alpha_{C} \Delta T$ , for cylinderNew radius will be $r_{C}^{'}=r_{C}+\Delta r_{C}$ $=r_{C}+r_{C} \alpha_{C} \Delta T=r_{C}(1+\alpha_{C} \Delta T)$ $=7.5 \times 10^{-2}[1+1.2 \times 10^{-5}(T-303)]$Similarly, new radius of piston, $r_{p}^{'}=7.492 \times 10^{-2}[1+1.6 \times 10^{-5}(T-303)]$Now, for fully fitted piston, $r_{C}=r_{p}$ $7.5 \times 10^{-2}[1+1.2 \times 10^{-5}(T-303)]$ $=7.492 \times 10^{-2}[1+1.6 \times 10^{-5}(T-303)]$ $\Rightarrow \;\; 1.0011+1.2013 \times 10^{-5}(T-303)$ $=1+1.6 \times 10^{-5}(T-303)$ $\Rightarrow 1.1 \times 10^{-3}=(T-303) \times 10^{-5}(1.6-1.2013)$ $=(T-303) \times 10^{-5} \times 0.3987$ $\Rightarrow T-303=275$ $\Rightarrow \;\; T=578\ \mathrm{K}=305^{\circ}\mathrm{C}$