Question

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 ₹ 16 per 100 cm2.

Answer 1

Inner radius (r) of hemispherical bowl r = $\frac{10.5}{2}$ cm = 5.25 cm
Surface area of hemispherical bowl = $2\pi r^2$
= 2 × $\frac{22}{7}$ × 5.25 cm × 5.25 cm
= 173.25 cm2

Cost of tinplating 100 cm2 area = Rs 16
Cost of tinplating 173.25 cm2 area
= ($₹\frac{16}{100}$) ×173.25 = 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72.