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Question -

The total surface area of a hollow metal cylinder, open at both ends of external radius 8 cm and height 10 cm is 338 π cm2. Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.



Answer -

Total surface area of a hollow metal cylinder = 338π cm2
Let R be the outer radius, r be inner radius and h be the height of thecylinder of the cylinder
∴ 2πRh +2πrh + 2πR2 – 2πr2 = 338π
R = 8 cm, h = 10 cm
⇒ 2πh (R+ r) + 2π(R2 – r2) = 338π
⇒Dividing by 2π , we get
⇒ h(R +r) + (R2 – r2) = 169
⇒ 10(8 +r) + (8 + r) (8 – r) = 169
⇒ 80 +10r + 64 – r2 = 169
⇒ 10r –r2 + 144 – 169 = 0
⇒ r2 –10r + 25 = 0
⇒ (r-5)2 =0
⇒ r = 5
∴Thickness of the metal = R – r = 8 – 5 = 3 cm

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