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Three particles A, Band C move in a circle of radius r = $\frac{1}{π}$ in anti-clockwise direction with speed 1 $m s^{- 1}$ , 2.5 $m s^{- 1}$ and 2 $m s^{- 1}$ , respectively. The initial positions of A, Band C are as shown in figure.

The ratio of distance travelled by B and C by the instant A, Band C meet for the first time is

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5 : 4

3 : 7

3 : 5

3 : 2

answer is B.

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Detailed Solution

$∵ \frac{d_{B}}{d_{C}} = \frac{v_{B} t}{v_{C} t} = \frac{v_{B}}{v_{C}} \Rightarrow \frac{d_{B}}{d_{C}} = \frac{25}{20} = \frac{5}{4}$

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