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