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.