A) \[T=2\pi \,\sqrt{\frac{Mh}{PA}}\]
B) \[T=2\pi \,\sqrt{\frac{Mh}{Ph}}\]
C) \[T=2\pi \,\sqrt{\frac{Mh}{PAh}}\]
D) \[T=2\pi \,\sqrt{MPhA}\]
Correct Answer: A
Solution :
[a] Let the piston be displaced through distance x towards left, then volume decreases, pressure increases. AP = increase in pressure and v = decrease in volume, for isothermal compression \[(P+\Delta p)\,(V-v)=PV\] \[(P+\Delta p)\,\,(Ah-Ax)=PV=PAh\] \[PAh-PAx+Ah\,\Delta p-\Delta p\,Ax=PAh\text{ }\] \[\text{ }\!\![\!\!\text{ }\Delta P\,Ax\,\,is\,negligible]\] \[-PAx+\Delta PAh=0\] \[\Delta p=\frac{Px}{h}\] \[\Delta F=\Delta pA\,=\frac{pAx}{h}\] Restoring force \[\Delta F\,=\frac{-PAx}{h}\] \[\Delta F\,=\frac{-PAx}{h}\] Compare it will \[a={{\omega }^{2}}x\] \[{{\omega }^{2}}=\frac{PA}{Mh}\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}\]You need to login to perform this action.
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