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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?
Solution
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
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