Test Index

Circular Motion

Section: Physics

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Question : 33 of 76

Marks: +1, -0

A particle is moving with constant speed in a circular path. When the particle turns by an angle

90^{\circ}

, the ratio of instantaneous velocity to its average velocity is

\pi: x \sqrt{2}

x

[6-Apr-2023 shift 1]

7
2
1
5

Solution: 👈: Video Solution

V_A = v \hat{j}

And

V_B = -v \hat{i}

Time to reach from

A

to

B = \frac{2\pi R}{4} \times \frac{1}{v} = \frac{\pi R}{2v}

Displacement from

A

to

B = R \sqrt{2}

Now, Average velocity from

A

to

B = \frac{\text{Displacement}}{\text{Time}} = \frac{R\sqrt{2}}{\frac{\pi R}{2v}} = \frac{2\sqrt{2}v}{\pi}

Instantaneous velocity at

B

is

-v \hat{i}

According to question,

\frac{\text{ instantaneous velocity }}{\text{ average velocity }} = \frac{\pi}{x\sqrt{2}}

\frac{v}{\frac{2\sqrt{2}v}{\pi}} = \frac{\pi}{x\sqrt{2}}

\Rightarrow \frac{\pi}{2\sqrt{2}} = \frac{\pi}{x\sqrt{2}}

\Rightarrow x = 2

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