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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 $= R - r$
$= 8 - 5$
$= 3\ cm$