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A hemispherical bowl is made of steel, $ 0.25 \mathrm{~cm} $ thick. The inner radius of the bowl is $ 5 \mathrm{~cm} $. Find the outer curved surface area of the bowl.
Given:
A hemispherical bowl is made of steel $0.25\ cm$ thick. The inside radius of the bowl is $5\ cm$.
To do:
We have to find the volume of steel used in making the bowl.
Solution:
The thickness of steel $= 0.25\ cm$
$=\frac{1}{4}\ cm$
Inside radius of the bowl $(r) = 5\ cm$
This implies,
Outside radius $(R) = 5 + 0.25$
$= 5.25\ cm$
Therefore,
Volume of the steel used $= \frac{1}{4} \pi(R^3-r^3)$
$=\frac{2}{3} \times \frac{22}{7} \times[(5.25)^{3}-(5.00)^{3}]$
$=\frac{44}{21}(144.703125-125.000000)$
$=\frac{44}{21} \times 19.703125$
$=41.28 \mathrm{~cm}^{3}$
The outer curved surface area of the bowl is $41.28 \mathrm{~cm}^{3}$.
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