Find the weight of a hollow sphere of metal having internal and external diameters as $20 c m$ and $22 c m$ , respectively if $1 {c m}^{3}$ of metal weighs $21 g$ .

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Solution

Volume of hollow sphere $= \frac{4}{3} π (r_{1}^{3} - r_{2}^{3}) = \frac{4}{3} \times \frac{22}{7} ({(11)}^{3} - {(12)}^{3})$
$= 1387.05 c m^{3}$
Weight of $1 c m^{3}$ of metal $= 21 g$
Weight of $1387.05 c m^{3}$ of metal $= 21 \times 1387.05 = 29128.05 g \approx 29.1 k g$

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Find the weight of a hollow sphere of metal having internal and external diameters as 20cm and 22cm, respectively if 1cm3 of metal weighs 21g.

Volume of hollow sphere=43π(r31−r32)=43×227((11)3−(12)3)=1387.05cm3Weight of 1cm3 of metal=21gWeight of 1387.05cm3 of metal=21×1387.05=29128.05g≈29.1kg

Find the weight of a hollow sphere of metal having internal and external diameters as $20 c m$ and $22 c m$ , respectively if $1 {c m}^{3}$ of metal weighs $21 g$ .

Volume of hollow sphere $= \frac{4}{3} π (r_{1}^{3} - r_{2}^{3}) = \frac{4}{3} \times \frac{22}{7} ({(11)}^{3} - {(12)}^{3})$
$= 1387.05 c m^{3}$
Weight of $1 c m^{3}$ of metal $= 21 g$
Weight of $1387.05 c m^{3}$ of metal $= 21 \times 1387.05 = 29128.05 g \approx 29.1 k g$