Heat and Thermodynamics 6 Question 15
19. While the piston is at a distance $2 L$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is
$(2007,4$ M)
(a) $\left(\frac{2 p _0 \pi R^{2}}{\pi R^{2} p _0+M g}\right)(2 L)$
(b) $\left(\frac{p _0 \pi R^{2}-M g}{\pi R^{2} p _0}\right)$ $(2 L)$
(c) $\left(\frac{p _0 \pi R^{2}+M g}{\pi R^{2} p _0}\right)(2 L)$
(d) $\left(\frac{p _0 \pi R^{2}}{\pi R^{2} p _0-M g}\right)(2 L)$
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Answer:
Correct Answer: 19. (d)
Solution:
- Let $p$ be the pressure in equilibrium.
Then, $\quad p A=p _0 A-M g$
$$ p=p _0-\frac{M g}{A}=p _0-\frac{M g}{\pi R^{2}} $$
Applying, $\quad p _1 V _1=p _2 V _2$
$$ \therefore \quad p _0(2 A L)=(p)\left(A L^{\prime}\right) $$
$\therefore L^{\prime}=\frac{2 p _0 L}{p}=\left(\frac{p _0}{p _0-\frac{M g}{\pi R^{2}}}\right)(2 L)$
$$ =\left(\frac{p _0 \pi R^{2}}{\pi R^{2} p _0-M g}\right)(2 L) $$
$\therefore$ Correct option is $(d)$.
