The total surface area of a hollow cylinder which is open on both the

1.	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» 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.

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»

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.