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Standard IX

Physics

Centripetal Force

A string of l...

Question

A string of length 1.5 m is tied to a stone of mass of 0.2 kg and is made to revolve at a speed of 1.2 m/s. The centripetal force experienced by the stone is.

0.173 N

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0.192 N

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0.160 N

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Solution

The correct option is B $0.192 N$
Given: Mass of the stone, $m = 0.2 k g$
Length of the string, $r = 1.5 m$
Speed of revolution, $v = 1.2 m / s$
Centripetal force, $F = \frac{m v^{2}}{r} = \frac{0.2 \times (1.2)^{2}}{1.5} = \frac{0.288}{1.5}$
$⟹ F = 0.192 N$

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A string of length 1.5 m is tied to a stone of mass of 0.2 kg and is made to revolve at a speed of 1.2 m/s. The centripetal force experienced by the stone is.

The correct option is B 0.192 NGiven: Mass of the stone, m=0.2 kg Length of the string, r=1.5 m Speed of revolution, v=1.2 m/s Centripetal force, F=mv2r=0.2×(1.2)21.5=0.2881.5 ⟹F=0.192 N

The correct option is B $0.192 N$
Given: Mass of the stone, $m = 0.2 k g$
Length of the string, $r = 1.5 m$
Speed of revolution, $v = 1.2 m / s$
Centripetal force, $F = \frac{m v^{2}}{r} = \frac{0.2 \times (1.2)^{2}}{1.5} = \frac{0.288}{1.5}$
$⟹ F = 0.192 N$