Question

find the thickness of the cylinder. the total surface area of the hollow cylinder which is open from both sides in 4609 sq cm, area of base ring is 115.5sq.cm and height 7 cm

Answer 1

hope this answer will help u

Answer 2

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.