# A hemispherical bowl made of brass has inner diameter 10.5 cm.

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. $6.93$