Ahemispherical bowl made of iron has inner diameter 84 cm. Find the cost of tin plating it on the inside at the rate of ₹21 per 100 cm² (take $\pi =\frac{22}{7}$) correct to two places of decimal.

₹2,328.48

We know that,

Curved Surface area of Hemisphere = 2πr²

Inner Diameter = 84 cm

Cost of tin plating it on the inside = Rs. 21 per cm².

Inner Diameter = 84 cm

then, Inner Radius = 42 cm

Curved Surface area of Hemisphere =  = 2 × \(\frac{22}{7}\) × 42² = 2 × 22 × 6 × 42 = 11088 cm²

The cost of tin-plating 100 cm²  = Rs. 21

The cost of tin-plating 1 cm²  = \(\frac{21}{100}\)

The cost of tin-plating 11088 cm² area of the bowl = \(\frac{21}{100}\) × 11088 = Rs. 2328.48