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# 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. $ 16 \mathrm{per} 100 \mathrm{~cm}^{2} $.

A hemispherical bowl made of brass has an inner diameter of $10.5\ cm$.

Rate of tin-plating is $Rs.\ 16$ per $100\ cm^2$.

To do:

We have to find the cost of tin-plating it on the inside.

Solution:

Inner diameter of the hemispherical bowl $= 10.5\ cm$

This implies,

Radius of the bowl $(r)=\frac{10.5}{2}$

$=5.25$

$=\frac{525}{100}$

$=\frac{21}{4} \mathrm{~cm}$

Therefore,

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

$=2 \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4}$

$=\frac{693}{4} \mathrm{~cm}^{2}$

Rate of tin-plating $= Rs.\ 16$ per $100 \mathrm{~cm}^{2}$

Total cost of tin-painting on the inside $=\frac{693 \times 16}{4 \times 100}$

$=Rs.\ \frac{2772}{100}$

$=Rs.\ 27.72$

The cost of tin-plating the bowl on the inside is $Rs.\ 27.72$.

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