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.
Answer & explanation
Correct answer: option 1
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