Question

The difference between outside and inside surfaces of a cylindrical metallic pipe $14$cm long is $44$ $c{m}^2$. If the pipe is made of $99c{m}^3$ of metal, find the outer and inner radii of the pipe.

Answer 1

Let, R be the external radius and r be the inner radius of the metallic pipe. $h=14$cm

Outer surface area$-$inner surface area$=44c{m}^2$

$\therefore 2\pi Rh-2\pi rh=44c{m}^2$

$\therefore R-r=\frac{44}{2\times \frac{22}{7}\times 14}=\frac{44\times 7}{2\times 22\times 14}$

$\therefore R-r=\frac{1}{2}$.

The volume of metal used $=99c{m}^3$

$\therefore$ External volume $-$ internal volume $=99c{m}^3$

$\therefore \pi R^{2}h-\pi r^{2}h=99c{m}^3$

$\therefore \pi h(R^2-r^2)=99c{m}^3$

$\therefore \frac{22}{7}\times 14(R+r)(R-r)=99c{m}^3$ $[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$ and $[\because R-r=1_{/2}]$

$22\times 2(R+r)\frac{1}{2}=99c{m}^3$

$\therefore R+r=\frac{99}{22}=\frac{9}{2}$

$R+r=\frac{9}{2}$

$\frac{R-r=\frac{1}{2}}{\frac{2R=\frac{10}{2}}{2R=5}}$

$R=\frac{5}{2}=2.5$cm

$\therefore$ External radius $=2.5$cm

$R+r=\frac{9}{2}$

$r+2.5=4.5$

$r=4.5-2.5=2$cm

Hence Internal radius $=2$cm