Question:
How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Solution:
Cube edge = 22 cm
Therefore volume of the cube $=(22)^{3}=10648 \mathrm{~cm}^{3}$
And,
Volume of each bullet $=4 / 3 \pi r^{3}$
$=4 / 3 \times 22 / 7 \times(1)^{3}$
= 4/3 × 22/7
$=88 / 21 \mathrm{~cm}^{3}$
Number of bullets $=\frac{\text { Volume of cube }}{\text { Volume of bullet }}=\frac{10648}{\frac{88}{21}}=2541$
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