AP EAPCET 04-Jul-2022 Shift 2 Paper

Initial temperature of the sink, $T_2=250 \text{ K}$ Initial efficiency of carnot engine,$\eta_1=25\%=0.25$ If $T_1$ be the temperature of the sink, then$\; \eta_1 = 1 - \frac{T_2}{T_1}$$\;0.25 = 1 - \frac{250}{T_1} \Rightarrow \frac{250}{T_1} = 0.75$$\; \Rightarrow \; \; T_1 = \frac{1000}{3} \text{ K}$When efficiency is increased by $50\%$ i.e $\; \; \eta_2 = 50\% = 0.5$Then if $T_2'$ be temperature of sink, then$\eta_2 = 1 - \frac{T_2'}{T_1}$ $\Rightarrow 0.5 = 1 - \frac{T_2'}{\frac{1000}{3}}$$\Rightarrow \frac{T_2'}{\frac{1000}{3}} = 1 - 0.5 = 0.5 = \frac{1}{2} .$$\Rightarrow T_2' = \frac{1}{2} \times \frac{1000}{3} = \frac{500}{3} \text{ K}$ Here we see that to increase the efficiency of carnot engine temperature of the sink should be decreased, but in question, asked about increasing temperature of sink to increase efficiency, which is not possible. Hence, no option is correct.