A long cylindrical pipe of radius 20 ,cm is c | vedclass

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(C) Let the initial pressure of the air inside be $p_0$ (atmospheric pressure) and initial volume be $V_i$. When a mass $m$ is attached to the piston,

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$0.04$

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(C) Let the initial pressure of the air inside be $p_0$ (atmospheric pressure) and initial volume be $V_i$. When a mass $m$ is attached to the piston,the piston moves down and the pressure of the gas becomes $p_f = p_0 - \frac{mg}{A}$,where $A$ is the cross-sectional area of the pipe.Since the temperature remains constant during this expansion,we use Boyle's Law: $p_i V_i = p_f V_f$.$p_0 V_i = (p_0 - \frac{mg}{A}) V_f$$\frac{V_f}{V_i} = \frac{p_0}{p_0 - \frac{mg}{A}} = \frac{1}{1 - \frac{mg}{p_0 A}}$Using the binomial approximation $(1-x)^{-1} \approx 1+x$ for small $x = \frac{mg}{p_0 A}$,we get $\frac{V_f}{V_i} \approx 1 + \frac{mg}{p_0 A}$.Thus,$\frac{\Delta V}{V_i} = \frac{V_f - V_i}{V_i} = \frac{mg}{p_0 A}$.When the gas is cooled from $T$ to $T-\Delta T$,it returns to its original volume $V_i$. For a constant pressure process (isobaric),$\frac{V}{T} = 	ext{constant}$,so $\frac{\Delta V}{V_f} = \frac{\Delta T}{T}$.Therefore,$\frac{\Delta T}{T} = \frac{\Delta V}{V_f} \approx \frac{\Delta V}{V_i} = \frac{mg}{p_0 A}$.Given $m = 50 \,kg$,$g = 10 \,m/s^2$,$p_0 = 10^5 \,Pa$,and $r = 0.2 \,m$ $(A = \pi r^2 = 3.14 	imes 0.04 = 0.1256 \,m^2)$:$\frac{\Delta T}{T} = \frac{50 	imes 10}{10^5 	imes 0.1256} = \frac{500}{12560} \approx 0.0398 \approx 0.04$.

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