A) \[T=2\pi \sqrt{\left( \frac{Mh}{PA} \right)}\]
B) \[T=2\pi \sqrt{\left( \frac{MA}{Ph} \right)}\]
C) \[T=2\pi \sqrt{\left( \frac{M}{PAh} \right)}\]
D) \[T=2\pi \sqrt{MPhA}\]
Correct Answer: A
Solution :
Let the piston be displaced through distance x towards left, then volume decreases, pressure increases. If \[\Delta P\] is increase in pressure and \[\Delta V\] is decrease in volume, then considering the process to take place gradually (i.e. isothermal) \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[\Rightarrow PV=(P+\Delta P)(V-\Delta V)\] \[\Rightarrow PV=PV+\Delta PV-P\Delta V-\Delta P\Delta V\] \[\Rightarrow \Delta P.V-P.\Delta V=0\] (neglecting \[\Delta P.\Delta V)\] \[\Delta P(Ah)=P(Ax)\]\[\Rightarrow \Delta P=\frac{P.x}{h}\] This excess pressure is responsible for providing the restoring force (F) to the piston of mass M. Hence \[F=\Delta P.A=\frac{PAx}{h}\] Comparing it with \[|F|=kx\Rightarrow k=M{{\omega }^{2}}=\frac{PA}{h}\] \[\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}\] Short trick : by checking the options dimensionally. Option (a) is correct.You need to login to perform this action.
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