Question:
A hemispherical bowl made of brass has inner diameter $10.5 \mathrm{~cm}$. Find the cost of tin plating it on the inside at the rate of Rs.4 per $100 \mathrm{~cm}^{2}$.
Solution:
Inner diameter of hemispherical bowl = 10.5cm
Radius = 10.5/2 = 5.25 cm
Surface area of hemispherical bowl $=2 \pi r^{2}$
$=2 \times 3.14 \times(5.25)^{2}$
$=173.25 \mathrm{~cm}^{2}$
Cost of tin plating $100 \mathrm{~cm}^{2}$ area $=$ Rs. 4
Cost of tin plating $173.25 \mathrm{~cm}^{2}$ area $=$ Rs. $\frac{4 \times 173.25}{100}=$ Rs. $6.93$
Thus, the cost of tin plating the inner side of hemispherical bowl is Rs. $6.93$
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