Question

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

Answer 1

Answer:

Let the radii of outer and inner surfaces be R and r.

(I) TSA of hollow cylinder :

TSA = Outer CSA + Inner CSA + 2(Area of circular base)

➳ 4620 = 2πRh + 2πrh + 2π(R² - r²)

➳ 4620 = 2πh(R + r) + 2 × 115.5

➳ 4620 = 2πh(R + r) + 231

➳ 4620 - 231 = 2πh(R + r)

➳ 4389 = 2πh(R + r)

➳ 4389 = 2 × 22/7 × 7 × (R + r)

➳ 4389 = 44 × (R + r)

➳ 4389/44 = (R + r)

➳ 399/4 = (R + r) ...........[Equation (i)]

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(II) Area of base ring :

Area of base ring = π(R² - r²)

➳ 115.5 = 22/7(R² - r²)

➳ 115.5 × 7 = 22(R² - r²)

➳ 808.5/22 = R² - r²

➳ 8085/22 = R² - r²

➳ 147/4 = (R + r) (R - r).......[Equation (ii)]

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Now, Substituting equation (I) in equation (II) we get,

➳ 147/4 = (R + r) (R - r)

➳ 147/4 = (399/4) (R - r)

➳ (R - r) = 399/147

➳ (R - r) = 7/19

➳ (R - R) = 0.36842 cm

Therefore, the thickness of the cylinder is 0.36842 cm.