A hemispherical bowl is packed in a cubical box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.

$= 2646 $ cm²

(c) Difference between the capacity of the bowl and the volume of the box:
Capacity of the bowl (inner volume) is given by
$ V_{\text{bowl}} = \frac{2}{3} \pi r^3 $
Substituting $r = 10$ cm and $\pi = 3.14$,
$ V_{\text{bowl}} = \frac{2}{3} \times 3.14 \times 10^3 $
$ = \frac{2}{3} \times 3.14 \times 1000 $
$ = \frac{6280}{3} = 2093.33 $ cm³

Volume of the cube is given by
$ V_{\text{cube}} = s^3 = 21^3 = 9261 $ cm³

Difference between the volumes:
$ \text{Difference} = V_{\text{cube}} - V_{\text{bowl}} $
$ = 9261 - 2093.33 = 7167.67 $ cm³

OR

(c) The area to be painted:
The inner surface of the bowl (inner curved surface area) is given by
$ A_{\text{inner}} = 2\pi r^2 $
Substituting $r = 10$ cm and $\pi = 3.14$,
$ A_{\text{inner}} = 2 \times 3.14 \times 10^2 $
$ = 2 \times 3.14 \times 100 = 628 $ cm²

The area of the thickness (outer curved surface - inner curved surface) is given by
$ A_{\text{thickness}} = 2\pi R^2 - 2\pi r^2 $
$ = 2\pi (R^2 - r^2) $
Substituting $R = 10.5$ cm and $r = 10$ cm,
$ A_{\text{thickness}} = 2 \times 3.14 \times (10.5^2 - 10^2) $
$ = 2 \times 3.14 \times (110.25 - 100) $
$ = 2 \times 3.14 \times 10.25 $
$ = 2 \times 32.185 = 64.37 $ cm²

Total area to be painted:
$ A_{\text{total}} = A_{\text{inner}} + A_{\text{thickness}} $
$ = 628 + 64.37 = 692.37 $ cm²