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A rectangular pipe having cross sectional area A. It is closed at one end and at its other end a block having same cross-section and mass ‘m’ is placed such that system is air tight. In equilibrium position of block the pressure and volume of air enclosed in pipe are P and V respectively. If block is displaced by small distance ‘x’ inward and released then find the time period of S.H.M. [Assume the walls are frictionless and compression of air is isothermal].

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$T = 2 π \sqrt{\frac{m V}{2 P A^{2}}}$

$T = \frac{π}{2} \sqrt{\frac{m V}{P A^{2}}}$

$T = 2 π \sqrt{\frac{P A^{2}}{m V}}$

$T = 2 π \sqrt{\frac{m V}{P A^{2}}}$

answer is D.

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Detailed Solution

$A d P = P A^{2} x a = \frac{P A^{2} x}{m V} = w^{2} x T = 2 π \sqrt{\frac{m V}{P A^{2}}}$

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