the total surface area of a hollow cylinder is open from both sides is 462026 area of a string is 15.5 square cm and height is 7 cm find the thickness of the cylinder
Share
the total surface area of a hollow cylinder is open from both sides is 462026 area of a string is 15.5 square cm and height is 7 cm find the thickness of the cylinder
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
Answer:
Solution:-
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
Answer.
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.