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(D) Given, time period $T = 40 \, s$.Total time $t = 2 \, 	ext{min} \, 20 \, s = 2 	imes 60 + 20 = 140 \, s$.Number of revolutions $n = \frac{t}{T}

Vedclass · Accepted Answer

$\frac{1}{11}$

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(D) Given, time period $T = 40 \, s$.Total time $t = 2 \, 	ext{min} \, 20 \, s = 2 	imes 60 + 20 = 140 \, s$.Number of revolutions $n = \frac{t}{T} = \frac{140}{40} = 3.5 = 3 \frac{1}{2}$ revolutions.Distance travelled in $3.5$ revolutions $= 3.5 	imes (2 \pi R) = 7 \pi R$.After $3.5$ revolutions, the particle is at the diametrically opposite point of the starting position. Thus, the displacement is equal to the diameter of the circle, which is $2R$.The ratio $\frac{|	ext{displacement}|}{	ext{distance}} = \frac{2R}{7 \pi R} = \frac{2}{7 \pi}$.Substituting $\pi = \frac{22}{7}$, we get $\frac{2}{7 	imes (22/7)} = \frac{2}{22} = \frac{1}{11}$.

$A$ particle is moving along a circle such that it completes one revolution in $40 \, s$. In $2 \, \text{minutes} \, 20 \, s$, the ratio $\frac{|\text{displacement}|}{\text{distance}}$ is:

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