Given:
- Inner radius of the hemispherical bowl, $r = 10$ cm
- Outer radius of the hemispherical bowl, $R = 10.5$ cm
- The bowl just fits in the cubical box, so its diameter equals the side of the cube
(a) Dimensions of the cubical box:
The side length of the cube is equal to the diameter of the bowl.
$ \text{Side of the cube} = 2R = 2 \times 10.5 $
$= 21 $ cm
Thus, the dimensions of the cubical box are $21 \times 21 \times 21$ cm.
(b) Total outer surface area of the box:
The surface area of a cube is given by
$ \text{Surface Area} = 6s^2 $
Substituting $s = 21$ cm,
$ \text{Surface Area} = 6 \times 21^2 $
$= 6 \times 441 $
$= 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