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.

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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 $41.28 \mathrm{~cm}^{3}$.

Updated on 10-Oct-2022 13:46:37