A) \[\frac{mg}{AK}\]
B) \[\frac{3mg}{AK}\]
C) \[\frac{mg}{3AK}\]
D) \[\frac{mg}{2AK}\]
Correct Answer: C
Solution :
[c] By definition. Bulk's modulus \[K=-\frac{dP}{\left( \frac{dV}{V} \right)}=-\frac{mg/A}{dV/V}\] \[V=\frac{4}{3}\pi {{R}^{3}}\Rightarrow \frac{dV}{V}=3\frac{dR}{R},\,\,\,K=\frac{mg}{3A\left( \frac{dR}{R} \right)}\] \[\Rightarrow \,\,\,\frac{dR}{R}=-\frac{mg}{3AK}\]You need to login to perform this action.
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