The thickness of hollow spherical shell is $2 c m$ . If its external radius is $8 c m$ find the volume of metal used in it.

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Solution

Given,External radius of hollow spherical shell $r_{1} = 8 c m$
Internal radius $r_{2} = 8 - 2 = 6$
$\begin{matrix} Volume of metal & = \frac{4}{3} π (r_{1}^{3} - r_{2}^{3}) = \frac{4}{3} \times \frac{22}{7} [(8)^{2} - (6)^{3}] \end{matrix}$
$\begin{matrix} = \frac{4}{3} \times \frac{22}{7} (512 - 216) = \frac{4 \times 22 \times 296}{21} = 1240.38 {c m}^{3} \end{matrix}$
Hence, volume of the metal used $= 1240.38 {c m}^{3}$

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The thickness of hollow spherical shell is 2cm. If its external radius is 8cm find the volume of metal used in it.

Given,External radius of hollow spherical shell r1=8cmInternal radius r2=8−2=6 Volume of metal =43π(r31−r32)=43×227[(8)2−(6)3]=43×227(512−216)=4×22×29621=1240.38cm3Hence, volume of the metal used =1240.38cm3

The thickness of hollow spherical shell is $2 c m$ . If its external radius is $8 c m$ find the volume of metal used in it.