Question

The total surface area of hollow cylinder, which is open from both sides, is $3575c{m}^2$; area of the base ring is $357.5c{m}^2$ and height is $14cm$. Find the thickness of the cylinder.

Answer 1

Let the inner radius be $r$

and outer radius be $R$

Base Ring $=\pi (R^2-r^2)=357.5c{m}^2$

$(R^2-r^2)=357.5÷22/7$

$(R+r)(R-r)=(3575\ast 7)/220$

$(R+r)(R-r)=113.75sqcm............\left(1\right)$

Total surface area of the cylinder $=3575sqcm$

Now, total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)

$=2\pi Rh+2\pi rh+2\pi (R^2-r^2)$

$=2\pi Rh+2\pi rh+2\ast 357.5=3575$

$=2\pi h(R+r)+2\times 357.5=3575$

$=2\pi h(R+r)+751=3575$

$=2\pi h(R+r)=3575-751$

$=2\times 22/7\times 14\times (R+r)=2824$

$=(R+r)=\frac{2824\ast 7}{44\ast 14}$

$=(R+r)=32$

Substituting the value of $(R+r)=32$ in equation (1), we get.

$(R+r)(R-r)=113.75$

$32(R-r)=113.75$

$R-r=\frac{113.75}{32}$

So, the thickness of the cylinder is $3.55cm$