Question

A particle is moving with constant speed in a circular path. When the particle turns by an angle \(90\degree\) , the ratio of instantaneous velocity to its average velocity is \(π : x\sqrt2\). The value of x will be -

• A. 7
• B. 2
• C. 1
• D. 5

Answer 1

The Correct Option is B

 

Velocity Analysis in Circular Motion

Given:

• \( V_A = v \hat{j} \)
• \( V_B = -v \hat{i} \)

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 = \( \frac{Displacement}{Time} = \frac{R\sqrt{2}}{\frac{\pi R}{2v}} = \frac{2\sqrt{2}v}{\pi} \)

Instantaneous velocity at B is \( -v \hat{i} \)

According to the question,

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

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

\( \frac{\pi}{2\sqrt{2}} = \frac{\pi}{x\sqrt{2}} \)

\( \implies x = 2 \) 

Conclusion:

The value of \( x \) is 2.

Answer 2

Circular Motion and Average Velocity

Step 1: Instantaneous Velocity

Let \( v \) be the constant speed of the particle. The instantaneous velocity at any point on the circular path is tangential to the path and has magnitude \( v \).

Step 2: Time for 90° Turn

The particle turns by an angle of 90° = \( \frac{\pi}{2} \) radians. Let \( R \) be the radius of the circular path.

The distance traveled by the particle is one-quarter of the circumference: \( s = \frac{1}{4} (2\pi R) = \frac{\pi R}{2} \).

Since the speed is constant, the time taken is \( t = \frac{s}{v} = \frac{\pi R}{2v} \).

Step 3: Displacement

When the particle turns by 90°, the displacement is the chord connecting the initial and final positions. Since it's a 90° turn, this forms a right-angled triangle with two sides equal to the radius \( R \). Using the Pythagorean theorem, the displacement magnitude is

\( d = \sqrt{R^2 + R^2} = R\sqrt{2} \)

Step 4: Average Velocity

The average velocity is the displacement divided by the time taken:

\( v_{avg} = \frac{d}{t} = \frac{R\sqrt{2}}{\frac{\pi R}{2v}} = \frac{2v\sqrt{2}}{\pi} \)

Step 5: Ratio of Instantaneous to Average Velocity

The ratio of the instantaneous velocity to the average velocity is:

\( \frac{v}{v_{avg}} = \frac{v}{\frac{2v\sqrt{2}}{\pi}} = \frac{\pi v}{2v\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \)

Given that this ratio is \( \pi : x\sqrt{2} \), we can write:

\( \frac{\pi}{2\sqrt{2}} = \frac{\pi}{x\sqrt{2}} \)

Therefore, \( x = 2 \).

Conclusion:

The value of \( x \) is 2 (Option 2).