A) First
B) Second
C) Third
D) All exert same pressure
Correct Answer: A
Solution :
\[\Pr essure=\frac{Force}{Area}\] \[p=\frac{F}{A}\,\,\,\,\Rightarrow \,\,p\propto \frac{1}{A}\] \[p\propto \frac{1}{{{r}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because \,\,A=4\pi {{r}^{2}})\] \[\because \] Sphere which has lesser area, will exert maximum pressure. As first sphere has small radius, so its area will be small. \[\therefore \] it will exert maximum pressure.
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