The total surface area of a hollow metal cylinder, open at both ends of external radius $8\ cm$ and height $10\ cm$ is $338 \pi\ cm^2$. Taking $r$ to be inner radius, obtain an equation in $r$ and use it to obtain the thickness of the metal in the cylinder.

 Given:

The total surface area of a hollow metal cylinder, open at both ends of external radius $8\ cm$ and height $10\ cm$ is $338 \pi\ cm^2$. 

To do:

We have to obtain an equation in $r$ and use it to obtain the thickness of the metal in the cylinder.

Solution:

Total surface area of a hollow metal cylinder $= 338\ pi\ cm^2$

Let $R$ be the outer radius, $r$ be the inner radius and $h$ be the height of the cylinder.

Therefore,

$2\pi Rh + 2\pi rh + 2\pi R^2 - 2\pi r^2 = 338\pi$

$2\pi h (R + r) + 2\pi (R^2 - r^2) = 338\pi$

Dividing by $2\pi$ on both sides, we get,

$h(R + r) + (R^2 - r^2) = 169$

$10(8 + r) + (8^2 - r^2)  = 169$

$80 + 10r + 64 - r^2 = 169$

$10r - r^2 + 144 - 169 = 0$

$r^2 - 10r + 25 = 0$

$(r-5)^2 = 0$

$r = 5$

This implies,

The thickness of the metal $= R - r$

$= 8 - 5$

$= 3\ cm$

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Updated on: 10-Oct-2022

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