Question

The total surface area of a hollow cylinder which is open from both sides is $4620$ sq. cm, area of base ring is $115.5$ sq. cm. and height $7$ cm. Find the thickness of the cylinder.

Answer 1

Given : Height of cylinder $\left(h\right)=7cm$
Total surface area of hollow cylinder $=4620c{m}^2$
Area of base ring $=115.5c{m}^2$

Let outer radius be ${}^{\prime }{R}^{\prime }$ cm, inner radius be ${}^{\prime }{r}^{\prime }$ cm

$\therefore$ Total surface area of hollow cylinder

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

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

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

Area of base $=\pi R^2-\pi r^2$

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

$\Rightarrow \frac{\text{Total Surface area}}{\text{Area of base}}=\frac{4620}{115.5}$

$\Rightarrow \frac{\left\{2\pi \right(R+r\left)\right(h+R-r\left)\right\}}{\left\{\pi \right(R+r\left)\right(R-r\left)\right\}}=\frac{4620}{115.5}$

$\Rightarrow \frac{2(h+R-r)}{R-r}=\frac{4620}{115.5}$

Consider thickness of cylinder to be ${}^{\prime }t=(R-r{)}^{\prime }$

$\Rightarrow \frac{2(h+t)}{t}=40$

$\Rightarrow 2h+2t=40t$

$\Rightarrow 2h=38t$

$\Rightarrow 2\left(7\right)=38t$

$\Rightarrow 14=38t$

$\Rightarrow t=\frac{14}{38}=\frac{7}{19}cm$

Hence, the thickness of cylinder is $\frac{7}{19}cm.$