A particle is moving with constant speed in a circular path. When the particle turns by an angle `90^@`, the ratio of instantaneous velocity to its average velocity is `pi : sqrt2 ` . The value of x will be

Options

A. `2`
B. `5`
C. `1`
D. `7`

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular path and understand the relationship between instantaneous velocity and average velocity when the particle turns by an angle of \(90^\circ\).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle moves in a circular path with a constant speed. When it turns by an angle of \(90^\circ\), it covers a quarter of the circular path.

2. **Defining Variables**:
- Let the radius of the circular path be \(r\).
- The distance covered by the particle when it turns by \(90^\circ\) is the length of the arc, which can be calculated as:
\[ \text{Arc length} = r \cdot \theta = r \cdot \frac{\pi}{2} \quad (\text{since } \theta = 90^\circ = \frac{\pi}{2} \text{ radians}) \]

3. **Calculating Average Velocity**:
- The average velocity (\(V_{avg}\)) is defined as the total displacement divided by the total time taken.
- The displacement when the particle moves \(90^\circ\) is the straight-line distance from the starting point to the endpoint, which forms the hypotenuse of a right triangle:
\[ \text{Displacement} = r\sqrt{2} \]
- If \(t\) is the time taken to move through the \(90^\circ\), then the average velocity is:
\[ V_{avg} = \frac{\text{Displacement}}{t} = \frac{r\sqrt{2}}{t} \]

4. **Calculating Instantaneous Velocity**:
- The instantaneous velocity (\(V_{inst}\)) at any point in circular motion is directed tangentially to the path. Since the particle is moving with constant speed, the magnitude of the instantaneous velocity is the same as the speed of the particle.
- Let this speed be \(v\). Thus, we have:
\[ V_{inst} = v \]

5. **Finding the Ratio**:
- According to the problem, the ratio of instantaneous velocity to average velocity is given as:
\[ \frac{V_{inst}}{V_{avg}} = \frac{v}{\frac{r\sqrt{2}}{t}} = \frac{vt}{r\sqrt{2}} \]
- We know from the problem statement that this ratio equals \(\frac{\pi}{\sqrt{2}}\):
\[ \frac{vt}{r\sqrt{2}} = \frac{\pi}{\sqrt{2}} \]

6. **Solving for \(x\)**:
- Rearranging the equation gives:
\[ vt = r\pi \]
- Since \(v = \frac{2\pi r}{T}\) (where \(T\) is the time period for a full circular motion), we can substitute \(v\) into the equation:
\[ \frac{2\pi r}{T} t = r\pi \]
- Dividing both sides by \(r\) (assuming \(r \neq 0\)):
\[ \frac{2\pi t}{T} = \pi \]
- Simplifying gives:
\[ 2t = T \quad \Rightarrow \quad t = \frac{T}{2} \]

7. **Final Calculation**:
- The value of \(x\) in the ratio \(\frac{v}{V_{avg}} = \frac{\pi}{x\sqrt{2}}\) is found by equating:
\[ x = 2 \]

Thus, the value of \(x\) is \(2\).