**Question:**

How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm?

**Solution:**

Given that, lots of spherical lead shots made from a solid rectangular lead piece.

∴ Number of spherical lead shots

$=\frac{\text { Volume of solid rectangular lead piece }}{\text { Volume of a spherical lead shot }}$ ..(i)

Also, given that diameter of a spherical lead shot $i . e$, , sphere $=4.2 \mathrm{~cm}$

$\therefore$ Radius of a spherical lead shot, $r=\frac{42}{2}=2.1 \mathrm{~cm}$ $\left[\because\right.$ radius $=\frac{1}{2}$ diameter $]$

So, volume of a spherical lead shot $i . e .$, sphere

$=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$

$=\frac{4 \times 22 \times 21 \times 21 \times 21}{3 \times 7 \times 1000}$

Now, length of rectangular lead piece, $l=66 \mathrm{~cm}$

Breadth of rectangular lead piece, $b=42 \mathrm{~cm}$

Height of rectangular lead piece, $h=21 \mathrm{~cm}$

$\therefore$ Volume of a solid rectangular lead piece $i . e .$, cuboid $=l \times b \times h=66 \times 42 \times 21$ From Eq. (i),

Number of spherical lead shots $=\frac{66 \times 42 \times 21}{4 \times 22 \times 21 \times 21 \times 21} \times 3 \times 7 \times 1000$

$=\frac{3 \times 22 \times 21 \times 2 \times 21 \times 21 \times 1000}{4 \times 22 \times 21 \times 21 \times 21}$

$=3 \times 2 \times 250$

$=6 \times 250=1500$

Hence, the required number of special lead shots is 1500.