The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm²; area of its base ring is 357.5 cm² and its height is 14 cm. Find the thickness of the cylinder.

Let external radius be R cm and internal radius be r cm.

Given,

Area of base of ring = 357.5 cm²

∴ π(R² - r²) = 357.5

$\Rightarrow \dfrac{22}{7} \times (R^2 - r^2) = 357.5 \\[1em] \Rightarrow (R^2 - r^2) = \dfrac{357.5 \times 7}{22} \\[1em] \Rightarrow (R^2 - r^2) = 113.75 \space ……….(1)$

Given,

Total surface area of hollow cylinder = 3575 cm².

$\Rightarrow 2πRh + 2πrh + 2π(R^2 - r^2) = 3575 \\[1em] \Rightarrow 2πh(R + r) + 2π \times 113.75 = 3575 \space ……….\text{[From (1)]} \\[1em] \Rightarrow 2 \times \dfrac{22}{7} \times 14 (R + r) + 2 \times \dfrac{22}{7} \times 113.75 = 3575 \\[1em] \Rightarrow 88(R + r) + 715 = 3575 \\[1em] \Rightarrow 88(R + r) = 3575 - 715 \\[1em] \Rightarrow 88(R + r) = 2860 \\[1em] \Rightarrow (R + r) = \dfrac{2860}{88} \\[1em] \Rightarrow (R + r) = 32.5 \space ……….(2)$

Dividing (1) by (2), we get :

$\Rightarrow \dfrac{R^2 - r^2}{R + r} = \dfrac{113.75}{32.5} \\[1em] \Rightarrow \dfrac{(R - r)(R + r)}{(R + r)} = 3.5 \\[1em] \Rightarrow (R - r) = 3.5 \text{ cm}.$

Hence, thickness of hollow cylinder = 3.5 cm.