A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 \({cm}^2\).

Inner Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{10.5}{2} = 5.25 cm\)

Surface area of sphere
= \(4 {\pi} {r}^2\)
= [\(4 × \frac{22}{7} × (5.25)^2\) \({cm}^2\)]
= [\(4 × \frac{22}{7} × 5.25 × 5.25\) \({cm}^2\)]
= ((88 × 5.25) \({cm}^2\)
= 173.25 \({cm}^2\)

Therefore, the surface area of a sphere having radius 5.25 cm is 173.25 \({cm}^2\).

Cost of tin-plating 100 \({cm}^{2}\) area = Rs. 16
Cost of tin-plating 173.25 \({cm}^{2}\) area = Rs.\(\frac{16 × 173.25}{100}\) = Rs. 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs. 27.72.