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A long container has air enclosed inside at room temperature and atmospheric pressure $(10^{5} p a) .$ It has a volume of $20, 000 c c .$ The area of cross section is $100 c m^{2}$ and force constant of spring is $k_{s p r i n g} = 1000 N / m .$ We push the right piston isothermally and slowly till it reaches the original position of the left piston which is movable. What is the final length of air column in cm. Assume that spring is initially relaxed.

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Detailed Solution

$P_{2} V_{2} {=P}_{1} V_{1} V_{1} =20000cc. P_{1} =105 Pa V_{2} =Ax P_{2} {=P}_{0} + \frac{k x}{A}$

$\Rightarrow 10^{5} \times 20000 \times 10^{- 6} = (10^{5} + \frac{1000 x}{100 \times 10^{- 4}}) 10^{- 4} \times 100 x \Rightarrow 2000 = 1000 [x (1+x)] x^{2} + x - 2 = 0 x = \frac{- 1 \pm \sqrt{1 + 8}}{2} = 1 m x = 100 c m$

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