Question

The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm²; area of its base ring is 357.5 cm² and its height is 14 cm. Find the thickness of the cylinder. 

Answer 1

Total surface area of a hollow cylinder = 3575 cm²

Area of the base ring = 357.5 cm²

Height = 14 cm

Let external radius = R and internal radius = r

Let thickness of the cylinder = d = (R – r)

Therefore, Total surface area = 2πRh + 2πrh + 2π(R² – r²) 

= 2πh(R + r) + 2π(R + r)(R – r)

= 2π(R + r)[h + R – r]

= 2π(R + r)(h + d)

= 2π(R + r)(14 + d)

But

2π(R + r)(14 + d) = 3575  ...(i)

And area of base = π(R² – r²) = 357.5

`=>` π(R + r)(R – r) = 357.5

`=>` π(R + r)d = 357.5   ...(ii)

Dividing (i) by (ii)

`(2pi(R + r) (14 + d))/(pi(R + r)d) = 3575/357.5`

`(2(14 + d))/d = 10`

28 + 2d = 10d

8d = 28

d = `28/8` = 3.5 cm

Hence, thickness of the cylinder = 3.5 cm