The total surface of a hollow metal cylinder, open at both ends of an external radius of 8cm and height 10 cm
The total surface of a hollow metal cylinder, open at both ends of an external radius of $8 \mathrm{~cm}$ and height $10 \mathrm{~cm}$ is $338 \mathrm{~cm}^{2}$. Take $r$ to be the inner radius, obtain an equation in $r$ and use it to obtain the thickness of the metal in the cylinder.
Given that
The external radius of the cylinder (R) = 8 cm
Height of the cylinder (h) = 10 cm
The total surface area of the hollow cylinder $(T . S . A)=338 \pi \mathrm{cm}^{2}$
As we know that,
$2 \pi r * h+2 \pi R * h+2 \pi R * 2-2 \pi r^{2}=338 \pi \mathrm{cm}^{2}$
⟹ h(r + R)+(R + r)(R − r) = 169
⟹ 10(8 + r) + (8 + r)(8 − r) = 169
$\Rightarrow 80+10 r+64-r^{2}=169$
$\Rightarrow r^{2}-10 r+25=0$
⟹ r = 5
R − r = 8 – 5 cm = 3 cm
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