A long cylindrical pipe of radius 20 ,cm is closed at its upper end and has an airtight piston of ne

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11.Thermodynamics

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A long cylindrical pipe of radius $20 \,cm$ is closed at its upper end and has an airtight piston of negligible mass as shown. When a $50 \,kg$ mass is attached to the other end of piston, it moves down by a distance $\Delta l$ before coming to equilibrium. Assuming air to be an ideal gas, $\Delta l / l$ (see figure) is close to $\left(g=10 \,m / s ^2\right.$, atmospheric pressure is $10^5 \,Pa$ ),

$0.01$

$0.02$

$0.04$

$0.09$

(KVPY-2017)

Solution

(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 $p$.

$\because\left(p_0-p\right) A=m g$

$\Rightarrow\left(p_0-\frac{p_0 l}{l+\Delta l}\right) A=m g \Rightarrow \frac{p_0}{m g} \frac{A}{\Delta l}=\frac{l}{\Delta l}+1$

$\Rightarrow 10^5 \times \pi \times\left(20 \times 10^{-2}\right)^2=\frac{l}{\Delta l}+1$

$\Rightarrow 50 \times 10$

$\Rightarrow \frac{22 \times 8}{\Delta}-1=\frac{l}{\Delta l}$

$\Rightarrow \frac{169}{7}$ or $\frac{\Delta l}{l} \approx 0.04$

Standard 11

Physics

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Solution

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