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

Edge of a cube = 22 cm

Volume of cube = side³ = 22³ = 10648 cm³.

Bullets are spherical in shape.

Radius of bullet, r = $\dfrac{\text{diameter}}{2} = \dfrac{2}{2} = 1 \text{cm.}$

Volume of each bullet = $\dfrac{4}{3}$ πr³

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 1^3 \\[1em] = \dfrac{88}{21} \text{ cm}^3.$

Let the number of bullets formed be n.

∴ Volume of cube = n × Volume of each bullet

$\Rightarrow 10648 = \text{n} \times \dfrac{88}{21} \\[1em] \Rightarrow \text{n} = \dfrac{10648 \times 21}{88} \\[1em] \Rightarrow \text{n} = \dfrac{223608}{88} \\[1em] \Rightarrow \text{n} = 2541$

Hence, 2541 bullets can be made out of a cube of lead.