Question

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

Answer 1

Given,

Total surface area of hollow cylinder = 4620 cm²

Area of base ring = 115.5 cm²

Height of cylinder = 7 cm

Let outer radius = R cm , inner radius = r cm

So,

Area of hollow cylinder = 2π(R² - r²) + 2πRh + 2πrh

= 2π(R + r)(R + r) + 2πh(R + r) = 2πh(R + r)(h + R - r)

Area of base = π(R² - r²)

∴ \(\frac{surface\,area}{area\,of\,base}\) = \(\frac{4620}{115.5}\)

= \(\frac{[2π(R+r)(h+R-r)}{[π(R+r)(R-r)]}\) = \(\frac{4620}{115.5}\)

= \(\frac{h+t}{t}\) = \(\frac{20}{1}\)

= h + t = 20t

= t = \(\frac{h}{19}\) = \(\frac{7}{19}\)

Thickness of cylinder = \(\frac{7}{19}\) cm