A Carnot engine has efficiency $\frac{1}{5}$. Efficiency becomes $\frac{1}{3}$ when the temperature of sink is decreased by $50 \mathrm{K}$. Temperature of the sink ( in $\mathrm{K}$ ) is 

The efficiency of the Carnot engine

$\mathrm{\eta }=\frac{\mathrm{Work}\mathrm{ }\mathrm{output}}{\mathrm{Heat}\mathrm{ }\mathrm{input}}=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{H}}}$

or $\mathrm{\eta }=\frac{{\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{Q}}_{\mathrm{H}}}=1-\frac{{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{Q}}_{\mathrm{H}}}$

Also we can show that

$\frac{{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{Q}}_{\mathrm{H}}}=\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$

$\therefore \mathrm{\eta }=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\mathrm{ }$

where ${\mathrm{T}}_{\mathrm{L}}$ is temperature of sink and ${\mathrm{T}}_{\mathrm{H}}$ is the temperature of hot reservoir.

According to question

$\mathrm{ }\frac{1}{5}=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}$ ...($\mathrm{i}$)

$\frac{1}{3}=1-\frac{{\mathrm{T}}_{\mathrm{L}}-50}{{\mathrm{T}}_{\mathrm{H}}}$ ...($\mathrm{ii}$)

From Eq. ($\mathrm{i}$)

$\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}=\frac{4}{5}$

$\Rightarrow {\mathrm{T}}_{\mathrm{H}}=\frac{5}{4}{\mathrm{T}}_{\mathrm{L}}$

Substituting value of ${\mathrm{T}}_{\mathrm{H}}$ in Eq. ($\mathrm{ii}$), we get

$\frac{1}{3}=1-\frac{{\mathrm{T}}_{\mathrm{L}}-50}{\frac{5}{4}{\mathrm{T}}_{\mathrm{L}}}$

$\mathrm{or} \frac{4({\mathrm{T}}_{\mathrm{L}}-50)}{5{\mathrm{T}}_{\mathrm{L}}}=\frac{2}{3}$

$\mathrm{or} \mathrm{ }{\mathrm{T}}_{\mathrm{L}}-50=\frac{2}{3}\times \frac{5}{4}{\mathrm{T}}_{\mathrm{L}}$

$\mathrm{or}\mathrm{ }{\mathrm{T}}_{\mathrm{L}}-\frac{5}{6}{\mathrm{T}}_{\mathrm{L}}=50$

$\therefore {\mathrm{T}}_{\mathrm{L}}=50\times 6=300\mathrm{ }\mathrm{K}$