SSC CGL 2 Sep 2016 Shif 2 Paper

Volume of the silver used in hollow hemispherical bowl
= $\frac{2}{3}\mathit{\pi }\left({\mathit{R}}^3-{\mathit{r}}^3\right)$
Where R = external radius
r = internal radius
= $\frac{2}{3}\mathit{\pi }\left(8^3-4^3\right)$ cu. cm
= $\frac{2}{3}\mathit{\pi }$ (512 - 64) cu. cm
= $\frac{2\mathit{\pi }}{3}$ × 488 cu. cm
∴ Volume of cone = $\frac{1}{3}\mathit{\pi }{\mathit{r}}_1^2\mathit{h}$
= $\frac{1}{3}\mathit{\pi }{8}^2$ × h
∴ $\frac{1}{3}\mathit{\pi }\times 8^2$ × h = $\frac{2\mathit{\pi }}{3}$ × 448
⇒ h = $\frac{2\times 448}{8\times 8}$ = 14 cm