The difference between outside and inside surface

We have to find the outer and inner radius of a hollow pipe.

Radius of inner pipe be $\left(r_{1}\right)$

Radius of outer cylinder be $\left(r_{2}\right)$

Length of the cylinder $(h)=14 \mathrm{~cm}$

Difference between the outer and the inner surface area is $44 \mathrm{~cm}^{2}$

So,

$2 \pi h\left(r_{2}-r_{1}\right)=44$

$2\left(\frac{22}{7}\right)(14)\left(r_{2}-r_{1}\right)=44$

So,

$\left(r_{2}-r_{1}\right)=\frac{1}{2}$...........(1)

So, volume of metal used is $99 \mathrm{~cm}^{3}$, so,

$\pi h\left(r_{2}^{2}-r_{1}^{2}\right)=99$

$\left(\frac{22}{7}\right)(14)\left(r_{2}-r_{1}\right)\left(r_{2}+r_{1}\right)=99$

Use equation (1) in the above to get,

$\left(\frac{22}{7}\right)(14)\left(\frac{1}{2}\right)\left(r_{2}+r_{1}\right)=99$

Therefore,

$\left(r_{2}+r_{1}\right)=\frac{9}{2} \ldots \ldots$(2)

Solve equation (1) and (2) to get,

$r_{2}=\frac{5}{2} \mathrm{~cm}$

$r_{1}=2 \mathrm{~cm}$