A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$. When the temperature of cold reservoir raised by x, its efficiency decreases to $\frac{1}{6}$. The value of x, if the temperature of hot reservoir is $99^\circ$C, will be :

Given $\eta=\frac{1}{3}$ <br/><br/>When $\mathrm{T}_2 $ raised by x ($\mathrm{T}_2 $ = $\left(\mathrm{T}_2+\mathrm{x}\right)$), its efficiency decreases to $\frac{1}{6}$<br/><br/>$\eta^{\prime}=\frac{1}{6}$ <br/><br/>Temperature of hot reservior $\left(\mathrm{T}_1\right)=99^{\circ} \mathrm{C}$ <br/><br/>$=99+273=372^{\circ} \mathrm{K}$ <br/><br/>As we know, efficiency of carnot engine <br/><br/>$$ \begin{aligned} & \eta=1-\frac{T_2}{T_1}=\frac{1}{3} .....(1) \\\\ & \eta^{\prime}=1-\frac{\left(T_2+x\right)}{T_1}=\frac{1}{6} .....(2) \\\\ & \eta^{\prime}=\frac{T_1-\left(T_2+x\right)}{T_1}=\frac{1}{6} \end{aligned} $$ <br/><br/>From equation (1) <br/><br/>$\frac{1}{3}=1-\frac{\mathrm{T}_2}{372}$ <br/><br/>$$ \begin{aligned} & \frac{1}{3}=\frac{372-\mathrm{T}_2}{372} \\\\ &\Rightarrow 372-\frac{372}{3}=\mathrm{T}_2 \\\\ &\Rightarrow \mathrm{~T}_2=248 \mathrm{~K} \end{aligned} $$ <br/><br/>By putting the value of $\mathrm{T}_2$ in equation (2) <br/><br/>$$ \begin{aligned} & \frac{T_1-\left(T_2+x\right)}{T_1}=\frac{1}{6} \\\\ &\Rightarrow \frac{372-(248+x)}{372}=\frac{1}{6} \\\\ &\Rightarrow 372-24-x=\frac{372}{6} \\\\ &\Rightarrow 124-x=62 \\\\ &\Rightarrow 124-62=x \\\\ &\Rightarrow x=62 \mathrm{~K} \end{aligned} $$