Calculate the percentage void space in closest packing of ‘n’ spheres of radius 1 unit, ‘n’ spheres of radius 0.414 units and ‘2n’ spheres of radius 0.225 units.

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answer is 19.

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Detailed Solution

ANS: 19
SOL: Percentage space occupied
$= \frac{\frac{n}{4} [4 \times \frac{4}{3} \bar{\land} {(1)}^{3} + 4 \times \frac{4}{3} \bar{\land} {(0.414)}^{3} + 4 \times 2 + \frac{4}{3} \bar{\land} {(0.225)}^{3}]}{\frac{n}{4} \times {(2 \sqrt{2})}^{3}} \times 100$

$∴ percentage void space = 100 - 81 = 19$

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