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