Question

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. 32 per 100 $c{m}^2$.

Answer 1

Here we will tin-plate the inside of the hemisphere So, We need to find the curved surface area of hemisphere

Radius of hemisphere = $\frac{r}{2}$ = $\frac{10.5}{2}$ cm = 5.25 cm

Curved surface area of the hemisphere = (2 $\times$ $\frac{22}{7}$ $\times$ 5.25 $\times$5.25) $c{m}^2$ =173.25 $c{m}^2$

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

Cost of tin-plating 1 $c{m}^2$ = Rs $\frac{1}{100}$ $\times$ $16$

Cost of tin-plating 173.25 $c{m}^2$ = Rs $\frac{1}{100}$ $\times$ $16$ $\times$ $173.25$ = Rs. $27.72$

∴ Cost of tin-plating the hemisphere is Rs $27.72$