Question

A hemispherical bowl made of brass has inner diameter $10.5cm$. Find the cost of tin-plating it on the inside at the rate of $Rs16per100c{m}^2$. (Assume $\mathrm{\pi }=\frac{22}{7}$)

Answer 1

Step 1: Find the inner curved surface area

The inner radius of the hemispherical bowl$=\frac{10.5}{2}=5.25cm$(Radius is half of the diameter)

To find the cost of tin-plating inside the bowl we have to find the inner curved surface area of the hemispherical bowl,

The inner curved surface area of the hemispherical bowl$=2{\mathrm{\pi r}}^2$

$2{\mathrm{\pi r}}^2=2\times \frac{22}{7}\times 5.25\times 5.25=173.25{\mathrm{cm}}^2$

Step 2: Find the Cost of tin-plating

Cost of tin-plating $100c{m}^2$ area = $Rs16$

So, the Cost of tin-plating $1c{m}^2$ area = $Rs\frac{16}{100}$

Cost of tin-plating $173.25c{m}^2$ area $=Rs\frac{16}{100}\times 173.25=Rs27.72$

Hence, the cost of tin-plating the inner side of the hemispherical bowl at the rate of $Rs16per100c{m}^2$ is $Rs27.72$.