A) \[\sqrt{\frac{Ag}{V}}\]
B) \[2\sqrt{\frac{Ag}{V}}\]
C) \[\sqrt{\frac{2Ag}{V}}\]
D) \[\frac{3Ag}{V}\]
Correct Answer: A
Solution :
Key Idea: Since, gas obeys Boyles law, so it is isothermal process. The bulk modulus of gas is given by \[B=-\frac{\Delta P}{\Delta V/V}\] But as gas obeys Boyles, so PV = constant (isothermal process). In isothermal process, isothermal bulk modulus of gas is equal to the pressure of the gas at that instant of time or \[B=P\] \[\therefore \] \[P=-\frac{\Delta P}{\Delta V/V}\] \[\Rightarrow \] \[\Delta P=-\frac{P}{V}\Delta V\] \[\Rightarrow \] \[\frac{F}{A}=-\frac{P}{V}Ax\] \[\Rightarrow \] \[F=-\frac{P{{A}^{2}}}{V}x\] This equation is similar to \[F=-kx\] where k is force constant of spring. So, \[k=\frac{P{{A}^{2}}}{V}\] Hence, angular frequency \[\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{P{{A}^{2}}}{mV}}=\sqrt{\frac{(mg)A}{mV}}\] \[\left( \because \,P=\frac{mg}{A} \right)\] \[\omega =\sqrt{\frac{Ag}{V}}\]You need to login to perform this action.
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