A) \[\frac{mg}{3\,AR}\]
B) \[\frac{mg}{A}\]
C) \[\frac{mg}{3AK}\]
D) \[\frac{mg}{AK}\]
Correct Answer: C
Solution :
For a spherical body \[V=\frac{4}{3}\pi {{R}^{3}}\] Differentiating and solving, we get \[\frac{\Delta R}{R}=\frac{1}{3}\frac{\Delta V}{V}\] We know that \[K=-\,V\frac{\Delta p}{\Delta V}\] Negative signs shows that when pressure is increased volume will decrease. i.e., \[\frac{\Delta V}{V}=\frac{\Delta p}{K}=\frac{mg}{AK}\] \[\left[ \because \Delta p=\frac{mg}{A} \right]\] Hence, \[\frac{\Delta R}{R}=\frac{1}{3}\frac{mg}{AK}\]You need to login to perform this action.
You will be redirected in
3 sec