Question

Solve the numerical example.

A large ball 2 m in radius is made up of a rope of square cross-section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?

Answer 1

Volume of ball = Volume enclosed by rope.

`4/3 pi` (radius)³ = Area of cross-section of rope × length of rope.

∴ length of rope l = `(4/3 pi"r"^3)/"A"`

Given: r = 2 m and 

Area = A = 4 × 4 = 16 m² = 16 × 10^-6 m²

∴ l = `(4 xx 3.142 xx 2^3)/(3 xx 16 xx 10^-6)`

= `(cancel4 xx 3.142 xx 2 xx 2 xx 2)/(3 xx cancel16 xx 10^(-6))`

= `(3.142 xx cancel8)/(3 xx cancel4 xx 10^(-6))`

= `(3.142 xx 2)/(3) xx 10^6`m

≈ 2 × 10⁶ m.

∴ Total length of rope to the nearest order of magnitude = 10⁶ m = 10³ km