Question

6.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm,
area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
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Answer 1

Answer:

Let the radii of outer and inner surfaces are R and r respectively.

∴ Area of the base ring = π(R² - r²)

⇒ 115.5 = π(R² - r²)

⇒ (R² - r²) = 115.5 ÷ 22/7

(R² - r²) = (115.5*7)/22

(R + r) (R - r) = (1155*7)/220

(R + r) (R - r) = 147/4 sq cm ............(1)

Total surface area of the cylinder = 4620 sq cm

Now, total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)

= 2πRh + 2πrh + 2π(R² - r²)

⇒ 2πRh + 2πrh + 2π(R² - r²) = 4620

⇒ 2πh (R + r) + (2 × 115.5) = 4620

⇒ 2πh (R + r) + 231 = 4620

⇒ 2πh (R + r) = 4620 - 231

⇒ 2 × 22/7 × 7 × (R + r) = 4389

⇒ (R + r) = 4389/44

⇒ (R + r) = 399/4

Substituting the value of (R + r) = 399/4 in equation (1), we get.

(R + r)(R - r) = 147/4

399/4 (R - r) = 147/4

R - r = 147/4 ÷ 399/4

R - r = (147/4) × (4/399)

R - r = 147/399

R - r = 7/19 cm

R - r = 0.368 cm

So, the thickness of the cylinder is 0.368 cm