Question

The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7 cm. Find the thickness of the cylinder.

Answer 1

Given: 

Total surface area of hollow cylinder = 4620 cm² 

Height of cylinder (h) = 7 cm

Area of base ring = 115.5 cm² 

Thickness of the cylinder Let ‘r₁’ and ‘r₂’ are the inner and outer radii of the hollow cylinder respectively.

Then, πr₂² – πr₁² = 115.5 …….(1) 

And, 2πr₁h + 2πr₂h+ 2(πr₂² – πr₁²) = 4620 

Or 2πh (r₁ + r₂ ) + 2 x 115.5 = 4620 

(Using equation (1) and h = 7 cm) 

or 2π7 (r₁ + r₂ ) = 4389 

or π (r₁ + r₂ ) = 313.5 ….(2) 

Again, from equation (1), 

πr₂² – πr₁² = 115.5 

or π(r₂ + r₁) (r₂ – r₁) = 115.5 

[using identity: a² – b² = (a – b)(a + b)] 

Using result of equation (2), 

313.5 (r₂ – r₁) = 115.5 

or r₂ – r₁ = \(\frac{7}{19}\) = 0.3684 

Therefore, thickness of the cylinder is \(\frac{7}{19}\) cm or 0.3684 cm.