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 $Rs.\ 4$ per $100\ cm^2$.
Given:
A hemispherical bowl made of brass has an inner diameter of $10.5\ cm$.
Rate of tin-plating is $Rs.\ 4$ 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,
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 $= Rs.\ 4$ per $100 \mathrm{~cm}^{2}$
Total cost of tin-painting on the inside $=\frac{693 \times 4}{4 \times 100}$
$=Rs.\ \frac{693}{100}$
$=Rs.\ 6.93$
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