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  • A particle is moving with constant speed in a circular path. When the particle turns by an angle 90o , the ratio of instantaneous velocity to its average velocity is <img alt

A particle is moving with constant speed in a circular path. When the particle turns by an angle 90o , the ratio of
instantaneous velocity to its average velocity is \pi:x\sqrt{2} The value of x will be -

Option: 1

7


Option: 2

2


Option: 3

1


Option: 4

5


Answers (1)

V_A=v \hat{\jmath}
And V_B=-v \hat{\imath}
Time to reach from \mathrm{A}$ to $\mathrm{B}=\frac{2 \pi R}{4} \times \frac{1}{v}=\frac{\pi R}{2 v}
Displacement from \mathrm{A} to \mathrm{B}=R \sqrt{2}
Now, Average velocity from \mathrm{A} to \mathrm{B}= \frac{\text { Displacement }}{\text { Time }}=\frac{R \sqrt{2}}{\frac{\pi R}{2 v}}=\frac{2 \sqrt{2} v}{\pi}
Instantaneous velocity at \mathrm{B} is -v \hat{i}
According to question, \frac{\text { instantaneous velocity }}{\text { average velocity }}=\frac{\pi}{x \sqrt{2}}
\begin{aligned} & \frac{v}{\frac{z \sqrt{2 v}}{\pi}}=\frac{\pi}{x \sqrt{2}} \\ & \Rightarrow \frac{\pi}{2 \sqrt{2}}=\frac{\pi}{x \sqrt{2}} \\ & \Rightarrow x=2 \end{aligned}

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Kshitij

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