A) \[\frac{mg}{AK}\]
B) \[\frac{2mg}{AK}\]
C) \[\frac{3mg}{AK}\]
D) \[\frac{mg}{3AK}\]
Correct Answer: D
Solution :
When mass m is placed on the piston, the excess pressure will be\[p=\frac{mg}{A}\]. This pressure acts equally from all the directions on the solid sphere. So, the radius of sphere decreases which in turn decreases the volume of sphere. \[\because \] \[V=\frac{4}{3}\pi {{r}^{3}}\] So \[\frac{\Delta V}{V}=3\frac{\Delta r}{r}\] Now, bulk modulus \[K=\frac{p}{\frac{\Delta V}{V}}\] \[\Rightarrow \] \[\frac{\Delta V}{V}=\frac{p}{K}=\frac{mg}{A}\times \frac{1}{K}\] \[\Rightarrow \] \[3\frac{\Delta r}{r}=\frac{mg}{AK}\] \[\therefore \] \[\frac{\Delta r}{r}=\frac{mg}{3AK}\]You need to login to perform this action.
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