Question

A hemispherical bowl made of brass has inner radius of 3.5 cm. Find the cost of tin-plating the inner part of the bowl as well as tin plating a plate that exactly fits as a lid for the bowl. Tin plating is done at the rate of ₹10 per 100 sq.cm.

Answer 1

Surface area that has to be plated = Curved surface area of the hemisphere bowl + surface area of the circular plate

$=2\pi r^2+\pi r^2$
$=2\times \frac{22}{7}\times {3.5}^2+\frac{22}{7}\times {3.5}^2$
$=3\times \frac{22}{7}\times {3.5}^2$
$=115.5{cm}^2$

Hence cost of tin plating
= Area x Rate
$=115.5\times \frac{10}{100}$
$=₹11.55$