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

We have,

Total surface area of hollow cylinder = 4620 cm²

Area of base ring = 115.5 cm²

Height of cylinder = 7 cm

Let outer radius be ‘R’ cm , inner radius be ‘r’ cm

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

= 2π(R+r)(R-r) + 2πh(R+r)

= 2π(R+r) (h+R-r)

Area of base = πR² – πr²

= π (R² – r²)

= π (R+r) (R-r)

Surface area / area of base = 4620/115.5

{2π(R+r) (h+R-r)} / { π (R+r) (R-r)} = 4620/115.5

2(h+R-r) / (R-r) = 4620/115.5

Let us consider (R-r) = t

2(h+t)/t = 40

2h + 2t = 40t

2h = 38t

2(7) = 38t

14 = 38t

t = 14/38

= 7/19 cm

∴ Thickness of cylinder is 7/19 cm.