The efficiency of carnot engine is $50\%$ and temperature of sink is $500\,K$ . If temperature of source is kept constant and its efficiency raised to $60\%$ , then the required temperature of the sink will be .... $K$
A$100$
B$600$
C$400$
D$500$
Medium
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C$400$
c $\mathrm{T}_{\sin \mathrm{k}}=500$
$\eta=50 \%$
$\mathrm{T}_{\text {source }}=1000 \mathrm{k}$
Now $\eta_{2}=60 \%$
$\frac{60}{100}=\frac{1000-\mathrm{T}_{2}}{1000}$
$\mathrm{T}_{2}=400 \mathrm{k}$
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