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

Answer:

Total surface area of cylinder = 3575 cm2

Area of the base ring = 357.5 cm2

Height = 14 cm

Let us consider

The inner radius = r

Outer radius = R

Area of the base ring

π(R2−r2) = 357.5 cm2

R2−r2 = 357.5/π

Value of π = 22/7 = 3.143

R2−r2 = 357.5/3.143

R2−r2 = 113.74 sq.cm

(R + r)(R – r) = 113.74 sq.cm…………………….(1)

Total surface area of a hollow cylinder = outer curved surface + inner curved surface area + 2(Area of the circular base)

3575 = 2πRh + 2πrh + 2π (R2−r2)

3575 = 2πRh+2πrh+ 2 × 357.5

3575 = 2πh(R+r) + 715

2πh(R+r) = 3575 – 715

2πh(R+r) = 2860

(R + r) = 2860/(2 × 3.143 × 14)

(R + r) = 2860/88.004

(R + r) = 32.498…………………………….(2)

Substitute the value of equation (2) in equation (1) we get

(32.498) × (R – r) = 113.74

(R – r) = 3.49 cm

The thickness of the cylinder is 3.49 cm.

Step-by-step explanation:

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