the total surface area of a hollow cylinder Which is open from both sides is 4620 square cm area of base ring is 115.5 square cm and height 7 cm find the thickness of the cylinder
Answers
base ring area = π(R² - r²)
R² - r² = 7*115.5/22 cm² = 73.5 cm²
Total surface area = 2 π h (r + R) = 2 *22/7*7 *(r + R) = 4620 cm²
r + R = 105 cm
R - r = 73.5 / 105 cm = 0.70 cm
Thickness = 7 mm
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.