Question

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.

 

Answer 1

\[\text{ Given } : \]
\[ \text{ Total surface area of the cylinder = 4620 {cm} }^2 \]
\[\text{ Area of the base ring = 115 . 5 {cm} }^2 \]
\[\text{ Height, h = 7 cm} \]
\[\text{ Let R be the radius of the outer ring and r be the radius of the inner ring } . \]
\[\text{ Area of the base ring } = \pi R^2 - \pi r^2 \]
\[115 . 5 = \pi\left( R^2 - r^2 \right)\]
\[ R^2 - r^2 = 115 . 5 \times \frac{7}{22} \]
\[(R + r)(R - r) = 36 . 75 . . . . . . . . . . . (i)\]
\[\text{ Total surface area = Inner curved surface area + Outer curved surface area + Area of bottom and top rings } \]
\[4620 = 2\pi rh + 2\pi Rh + 2 \times 115 . 5\]
\[2\pi h(R + r) = 4620 - 231\]
\[R + r = \frac{4389 \times 7}{2 \times 22 \times 7}\]
\[R + r = \frac{399}{4} . . . . . . . . . . . (ii) \]
\[\text{ Substituting the value of R + r from the equation (ii) in } (i): \]
\[\frac{399}{4}(R-r) = 36 . 75\]
\[(R - r) = 36 . 75 \times \frac{4}{399} = 0 . 368 \text{ cm } \]
\[ \therefore \text{ Thickness of the cylinder = (R - r) = 0 . 368 cm } \]