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Question

(c)

Initially pressure inside the cylinder is atmospheric pressure $p_0$. When mass $m$ is attached to piston and it comes down by a distance $\Del

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

(c)

Initially pressure inside the cylinder is atmospheric pressure $p_0$. When mass $m$ is attached to piston and it comes down by a distance $\Delta l$, let pressure is $p$.

Then, in equilibrium,

$p_0 V_0=p V \Rightarrow p_0(A)(l)=p A(l+\Delta l)$

So, final pressure will be

$p=\frac{p_0 A l}{A(l+\Delta l)}=\frac{p_0 l}{(l+\Delta l)}$

In equilibrium, weight of mass $m$ is balanced by force of suction due to reduced pressure $p$.

$\because\left(p_0-pight) A=m g$

$\Rightarrow\left(p_0-\frac{p_0 l}{l+\Delta l}ight) A=m g \Rightarrow \frac{p_0}{m g} \frac{A}{\Delta l}=\frac{l}{\Delta l}+1$

$\Rightarrow 10^5 	imes \pi 	imes\left(20 	imes 10^{-2}ight)^2=\frac{l}{\Delta l}+1$

$\Rightarrow 50 	imes 10$

$\Rightarrow \frac{22 	imes 8}{\Delta}-1=\frac{l}{\Delta l}$

$\Rightarrow \frac{169}{7}$ or $\frac{\Delta l}{l} \approx 0.04$

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