A) \[\frac{Mg}{AK}\]
B) \[\frac{Mg}{3AK}\]
C) \[\frac{3Mg}{AK}\]
D) \[\frac{Mg}{2AK}\]
Correct Answer: B
Solution :
Change in pressure due to placing of mass on piston is, \[\Delta p=\frac{Mg}{A}\] From bulk modulus definition\[K=\frac{-dp}{dV/V}\] \[\Rightarrow \] \[\left| \frac{dV}{V} \right|=\frac{\Delta p}{K}=\frac{Mg}{AK}\] From \[V=\frac{4}{3}\pi {{r}^{3}}\] \[\frac{dV}{V}=\frac{3dR}{R}\] \[\Rightarrow \] \[\frac{dR}{R}=\frac{1}{3}\frac{dV}{V}=\frac{Mg}{3AK}\]You need to login to perform this action.
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