⇒Thermodynamics

Correct Option is (B)

Efficiency of a carnot engine is $\eta =1-\frac{T_C}{T_H}$

For case (i),

${\mathrm{T}}_{\mathrm{C}}=100\mathrm{K}$ and ${\mathrm{T}}_{\mathrm{H}}=600\mathrm{K}$

$\therefore {\eta }_1=1-\frac{100}{600}=\frac{5}{6}$

For case (ii),

${\mathrm{T}}_{\mathrm{C}}=\mathrm{T}\mathrm{K}\text{ and }{\mathrm{T}}_{\mathrm{H}}=960\mathrm{K}$

$\therefore {\eta }_2=1-\frac{\mathrm{T}}{960}$

Given ${\eta }_1={\eta }_2$

$\begin{array}{rl}\therefore \frac{5}{6}& =1-\frac{\mathrm{T}}{960}\\ \frac{5}{6}& =\frac{960-\mathrm{T}}{960}\end{array}$

Solving for $\mathrm{T}$, we get $\mathrm{T}=160\mathrm{K}$

Correct Option is (B)

$\begin{array}{rl}& T_H=600\mathrm{k}\\ & \eta =1-\frac{T_C}{T_H}\\ & \text{ but, }\eta =\frac{1}{2}....(\text{Given}:\eta =50\mathrm{\%})\\ & \Rightarrow \frac{1}{2}=1-\frac{T_C}{600}\\ & \therefore T_C=300K\end{array}$

With ${\mathrm{T}}_{\mathrm{C}}=300\mathrm{K}$, the efficiency is increased to $70\mathrm{\%}$

$\therefore$ New temperature of the source will be ${\mathrm{T}}_{{\mathrm{H}}_{\text{new }}}$

$\begin{array}{l}\therefore \eta =1-\frac{300}{{\mathrm{T}}_{{\mathrm{H}}_{\text{new }}}}\\ \frac{300}{{\mathrm{T}}_{{\mathrm{H}}_{\text{new }}}}=1-\frac{7}{10}\dots .(\because \eta =70\mathrm{\%})\\ \therefore {\mathrm{T}}_{{\mathrm{H}}_{\text{new }}}=\frac{3000}{3}=1000\mathrm{K}\end{array}$