Question

The difference between the outside and inside surface of a cylindrical metallic pipe 14 cm long is $44c{m}^2$. If the pipe is made of 99 $c{m}^3$ of metal, find the outer and inner radii of the pipe. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

Let, external radius = R cm and internal radius = r cm

Then, outside surface
$=2\pi Rh=(2\times \frac{22}{7}\times R\times 14)c{m}^2=\left(88R\right)c{m}^2$.

Inside surface
$=2\pi rh=(2\times \frac{22}{7}\times r\times 14)c{m}^2=\left(88r\right)c{m}^2$.

Given : Outside surface - Inside surface = 44 $c{m}^2$
$\therefore (88R-88r)=44$

$\Rightarrow (R-r)=\frac{44}{88}=\frac{1}{2}$

$\Rightarrow (R-r)=\frac{1}{2}\dots \dots \left(i\right)$

External volume
$=\pi R^{2}h=(\frac{22}{7}\times R^2\times 14)c{m}^3=\left(44R^2\right)c{m}^3$.

Internal volume
$=\pi r^{2}h=(\frac{22}{7}\times r^2\times 14)c{m}^3=\left(44r^2\right)c{m}^3$.

Given : External volume - Internal volume = 99$c{m}^3$
$\therefore (44R^2-44r^2)=99$

$\Rightarrow (R^2-r^2)=\frac{99}{44}$

$\Rightarrow (R^2-r^2)=\frac{9}{4}\dots \dots \left(ii\right)$

On dividing (ii) by (i), we get:
$(R+r)=(\frac{9}{4}\times \frac{2}{1})\Rightarrow (R+r)=\frac{9}{2}\dots \dots \left(iii\right)$

Solving (i) and (iii) , we get, R = 2.5 and r =2.

Hence, outer radius = 2.5 cm and inner radius = 2cm.