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.

Given:

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 $7\ cm$. 

To do:

We have to find the thickness of the cylinder.

Solution:

Total surface area of a hollow cylinder open from both sides $= 4620\ cm^2$

Area of the base of ring $= 115.5\ cm^2$

Height $(h) = 7\ cm$

Let the outer radius be $R$ and the inner radius be $r$.

This implies,

Total surface area $=2 \pi \mathrm{R} h+2 \pi r h+2 (\mathrm{R}^{2}-r^{2})$

$=4620$

$\Rightarrow 2\pi h(\mathrm{R}+r)+2 \pi(\mathrm{R}^{2}-r^{2})=4620$

$\Rightarrow  \pi h(\mathrm{R}+r)+2 \times 115.5=4620$                  $From} $i)]

$\Rightarrow & 2 \times \frac{22}{7} \times 7(\mathrm{R}+r)+231=4620$

$\Rightarrow 44(\mathrm{R}+r)=4620-231=4389$

$\mathrm{R}+r=\frac{4389}{44}$

$=\frac{399}{4}$$..................(i)

Dividing (i) by (ii)

$\$rac{\pi\left(\mathrm{R}^{2}-r^{2}\right)}{\mathrm{R}+r}=\frac{115.5 \times 4}{399}$

$\frac{22}{7}(\mathrm{R}-r)=\frac{1155 \times 4}{10 \times 399}$

$\mathrm{R}-r=\frac{1155 \times 4 \times 7}{10 \times 399 \times 22}=\frac{21}{57}=\frac{7}{19} $

$\mathrm{R}-r=\frac{7}{19}$

Thickness of the $cylinder =\frac{7}{19} \mathrm{~cm}$

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Updated on: 10-Oct-2022

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