the total surface area of a hollow cylinder which is open from both sides is 4620 sq.cm., area of base ring is 115.5 sq cm and height 7 cm .find the thickness of the cylinder
Answers
Let the radii of outer and inner surfaces are R and r respectively.
∴ Area of the base ring = π(R² - r²)
⇒ 115.5 = π(R² - r²)
⇒ (R² - r²) = 115.5 ÷ 22/7
(R² - r²) = (115.5*7)/22
(R + r) (R - r) = (1155*7)/220
(R + r) (R - r) = 147/4 sq cm ............(1)
Total surface area of the cylinder = 4620 sq cm
Now, total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)
= 2πRh + 2πrh + 2π(R² - r²)
⇒ 2πRh + 2πrh + 2π(R² - r²) = 4620
⇒ 2πh (R + r) + (2 × 115.5) = 4620
⇒ 2πh (R + r) + 231 = 4620
⇒ 2πh (R + r) = 4620 - 231
⇒ 2 × 22/7 × 7 × (R + r) = 4389
⇒ (R + r) = 4389/44
⇒ (R + r) = 399/4
Substituting the value of (R + r) = 399/4 in equation (1), we get.
(R + r)(R - r) = 147/4
399/4 (R - r) = 147/4
R - r = 147/4 ÷ 399/4
R - r = (147/4) × (4/399)
R - r = 147/399
R - r = 7/19 cm
R - r = 0.368 cm
So, the thickness of the cylinder is 0.368 cm
I hope this is correct and if it helped you please mark it brainliest
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)]
_____________________
(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)]
____________________
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.