Question

Question 5
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 $Rs.16per100c{m}^2.[Assume\pi =\frac{22}{7}]$

Answer 1

Inner radius (r) of the hemispherical bowl $=\left(\frac{10.5}{2}\right)cm=5.25cm$

Surface area of the hemispherical bowl $=2\pi r^2$

$=[2\times \frac{22}{7}\times (5.25{)}^2]c{m}^2=173.25c{m}^2$

$\because$ Cost of tin-plating $100c{m}^2$ area = Rs. 16

$\therefore$ Cost of tin-plating of $173.25c{m}^{2}area=\left(\frac{16\times 173.25}{100}\right)$
= Rs. 27.72