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

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(C) Initially,the pressure inside the cylinder is the atmospheric pressure $p_0$. When a mass $m$ is attached to the piston and it moves down by a dis

Vedclass · Accepted Answer

$0.04$

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(C) Initially,the pressure inside the cylinder is the atmospheric pressure $p_0$. When a mass $m$ is attached to the piston and it moves down by a distance $\Delta l$,let the new pressure be $p$.Since the process is isothermal (assuming temperature remains constant),we have $p_0 V_0 = p V$.$p_0 (A l) = p A (l + \Delta l)$$p = p_0 \frac{l}{l + \Delta l}$In equilibrium,the downward force due to the mass $m$ is balanced by the upward force due to the pressure difference between the atmosphere and the gas inside.$(p_0 - p) A = m g$Substituting $p$:$(p_0 - p_0 \frac{l}{l + \Delta l}) A = m g$$p_0 A (1 - \frac{l}{l + \Delta l}) = m g$$p_0 A (\frac{\Delta l}{l + \Delta l}) = m g$Given $p_0 = 10^5 \,Pa$,$m = 50 \,kg$,$g = 10 \,m/s^2$,$r = 0.2 \,m$,$A = \pi r^2 = \pi (0.2)^2 = 0.04 \pi \,m^2$.$10^5 	imes 0.04 \pi 	imes \frac{\Delta l}{l + \Delta l} = 50 	imes 10 = 500$$4000 \pi 	imes \frac{\Delta l}{l + \Delta l} = 500$$\frac{\Delta l}{l + \Delta l} = \frac{500}{4000 \pi} = \frac{1}{8 \pi} \approx \frac{1}{8 	imes 3.14} \approx \frac{1}{25.12} \approx 0.0398$Since $\frac{\Delta l}{l + \Delta l} \approx 0.04$,we have $\frac{\Delta l}{l} \approx 0.04$ (assuming $\Delta l \ll l$).Thus,the correct option is $C$.

Column-$I$	Column-$II$
$(a)$ Top graph	$(i)$ Constant pressure
$(b)$ Bottom graph	$(ii)$ Constant volume
	$(iii)$ For ideal gas

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Column-$I$ represents a graph and Column-$II$ represents a physical quantity or condition that is constant for that graph. Match them correctly:
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$(a)$ Top graph $(i)$ Constant pressure
$(b)$ Bottom graph $(ii)$ Constant volume
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