The total surface area of a hollow cylinder which is open from both sides is 4620sq. cm, area of base ring is 115.5sq. cm and height 7 cm. Find the thickness of the cylinder.(with explanation please)
Answers
curved surface area + 2(area of base rings) = 2h(R+r) + 2(R2 -r2).....................1 also, = 2(R +r)(h + R -r)...........................2
total surface area of hollow cylinder = 4620 sq cm
so,
2h(R+r) + 2(R2 -r2) = 4620
area of base rings = 115.5 sq cm
so,
2h(R+r) = 4620 - 2(115.5)
2(R+r) = 4389/7 {here h = 7cm} =627....................3
now from eq 2
2(R +r)(h + R -r) = 4620
(h + R -r)=4620/627 {from eq 3}
R -r = 7.36842 - 7
R -r = .36842 cm
so thickness of given cylinder = .36842 cm
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.