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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,

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प्रश्न

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?

संख्यात्मक
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उत्तर

Given: r = 2 m

Area = A = 4 × 4 = 16 mm2 = 16 × 10−6 m2

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 pir^3)/A`

∴ 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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अध्याय 1: Units and Measurements - Exercises [पृष्ठ १४]

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बालभारती Physics [English] Standard 11 Maharashtra State Board
अध्याय 1 Units and Measurements
Exercises | Q 3. iv) | पृष्ठ १४

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