
Initial pressure in cylinder is atmospheric pressure $p_0$.
When mass $m$ is attached to piston, then pressure $=p_0-\frac{m g}{A}$.
As, temperature remains constant during expansion
$\Rightarrow p_i V_i=p_f V_f$
$\Rightarrow p_0 V_i=\left(p_0-\frac{m g}{A}\right) \cdot V_f$
$\Rightarrow \frac{V_f}{V_i}=\frac{p_0}{\left(p_0-\frac{m g}{A}\right)}$
$\frac{V_i}{V_f}=1-\frac{m g}{p_0 A} \Rightarrow \frac{m g}{p_0 A}=1-\frac{V_i}{V_f}$
$\Rightarrow \frac{m g}{p_0 A}=V_f-V_i=\frac{\Delta V}{V_f}$
Now, when temperature is reduced by $\Delta T$, the volume of gas again contracts to its original volume.
$\Rightarrow \frac{V}{T}=$ constant
or $\quad \frac{\Delta V}{V}=\frac{\Delta T}{T}$
$\Rightarrow \frac{\Delta T}{T}=\frac{\Delta V}{V_f}=\frac{m g}{p_0 A}$
$\Rightarrow \frac{\Delta T}{T}=\frac{m g}{p_0 A}$
$=\frac{50 \times 10}{10^5 \times 3.14 \times(0.2)^2}$
$=\frac{5}{3.14 \times 4} \times \frac{10^2}{10^5 \times 10^{-2}}$
$=0.4 \times 10^{-1}=0.04$


(image)
$(A)$ Process $I$ is an isochoric process $(B)$ In process $II$, gas absorbs heat
$(C)$ In process $IV$, gas releases heat $(D)$ Processes $I$ and $III$ are $not$ isobaric
