How many bullets can be made out of a cube of lead, whose edge measures $22\ cm$, each bullet being $2\ cm$ in diameter?

 Given:

The edge of a lead cube is $22\ cm$.

Diameter of each bullet is $2\ cm$.

To do:

We have to find the number of bullets that can be made out of the cube of lead.

Solution:

Edge of the lead cube $(k) = 22\ cm$

This implies,

Volume of the lead cube $= k^3$

$= (22)^3\ cm^3$

$= 10648\ cm^3$

Diameter of the bullet $= 2\ cm$

This implies,

Radius of the bullet $(r)=\frac{2}{2}$

$=1 \mathrm{~cm}$

Therefore,

Volume of each bullet $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 1^3$

$=\frac{88}{21} \mathrm{~cm}^{3}$

Number of bullets that can be made $=\frac{\text { Volume of the lead cube }}{\text { Volume of each bullet }}$

$=\frac{10648 \times 21}{88}$

$=2541$

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Updated on: 10-Oct-2022

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