3 Tutor System
Starting just at 265/hour

# 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.