**Question:**

A cylinder is open at both ends and is made of 1.5-cm-thick metal. Its external diameter is 12 cm and height is 84 cm. What is the volume of metal used in making the cylinder? Also, find the weight of the cylinder if 1 cm3 of the metal weighs 7.5 g.

**Solution:**

Thickness of the cylinder $=1.5 \mathrm{~cm}$

External diameter $=12 \mathrm{~cm}$

i.e., radius $=6 \mathrm{~cm}$

also, internal radius $=4.5 \mathrm{~cm}$

Height $=84 \mathrm{~cm}$

Now, we have the following:

Total volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 6 \times 6 \times 84=9504 \mathrm{~cm}^{3}$

Inner volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 4.5 \times 4.5 \times 84=5346 \mathrm{~cm}^{3}$

Now, volume of the metal $=\operatorname{total}$ volume $-$ inner volume $=9504-5346=4158 \mathrm{~cm}^{3}$\

$\therefore$ Weight of iron $=4158 \times 7.5=31185 \mathrm{~g}=31.185 \mathrm{~kg}\left[\right.$ Given: $\left.1 \mathrm{~cm}^{3}=7.5 \mathrm{~g}\right]$