A) \[\frac{1}{2\pi }\,\frac{A\gamma {{P}_{0}}}{{{V}_{0}}M}\]
B) \[\frac{1}{2\pi }\,\frac{{{V}_{0}}M{{P}_{0}}}{{{A}^{2}}\gamma }\]
C) \[\frac{1}{2\pi }\,\sqrt{\frac{{{A}^{2}}\gamma {{P}_{0}}}{M{{V}_{0}}}}\]
D) \[\frac{1}{2\pi }\,\sqrt{\frac{M{{V}_{0}}}{A\gamma {{P}_{0}}}}\]
Correct Answer: C
Solution :
\[\frac{Mg}{A}={{P}_{0}}\] |
\[Mg={{P}_{0}}A\] ...(1) |
\[{{P}_{0}}{{V}_{0}}^{\gamma }=P{{V}^{\gamma }}\] |
\[{{P}_{0}}A{{x}_{0}}^{\gamma }=PA{{({{x}_{0}}-x)}^{\gamma }}\] |
\[P=\frac{{{P}_{0}}x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}}\] |
Let piston is displaced by distance . |
\[Mg-\left( \frac{{{P}_{0}}x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}} \right)A={{F}_{restoring}}\] \[{{P}_{0}}A\left( 1-\frac{x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}} \right)={{F}_{restoring}}\] |
\[F=-\frac{\gamma {{P}_{0}}Ax}{{{x}_{0}}}\] |
\[\therefore \] Frequency with which piston executes SHM. |
\[f=\frac{1}{2\pi }\sqrt{\frac{\gamma {{P}_{0}}A}{{{x}_{0}}M}}=\frac{1}{2\pi }\sqrt{\frac{\gamma {{P}_{0}}{{A}^{2}}}{M{{V}_{0}}}}\] |
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