At high altitude , a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of `10 m//s`. Time taken by the two radius vectors connecting of explosion to fragments to make `90^(@)` is

Options

A. 0.5 s
B. 4 s
C. 2 s
D. 1 s

Explanation

To solve the problem step by step, we will analyze the situation of the explosion of the body into two equal fragments and how their velocities relate to the time taken for the radius vectors to make a 90-degree angle.

### Step-by-Step Solution:

1. **Understanding the Explosion**:
- A body at rest explodes into two equal fragments. After the explosion, one fragment receives a horizontal velocity of \(10 \, \text{m/s}\).

2. **Velocity of the Fragments**:
- Since the body was initially at rest, the total momentum before the explosion is zero. Therefore, the momentum after the explosion must also be zero.
- If one fragment moves horizontally with a velocity of \(10 \, \text{m/s}\), the other fragment must have a velocity that balances this momentum. Since they are equal fragments, we can assume the second fragment moves in a direction that results in a net momentum of zero.

3. **Direction of the Second Fragment**:
- Let’s assume the second fragment moves at an angle of \(45^\circ\) to the horizontal. The horizontal component of its velocity must be equal and opposite to that of the first fragment.
- Therefore, if the first fragment moves with a velocity \(v_1 = 10 \, \text{m/s}\) horizontally, the second fragment must have a horizontal component of velocity \(v_{2x} = -10 \, \text{m/s}\).

4. **Finding the Vertical Component**:
- Since the second fragment is moving at \(45^\circ\), we can find its vertical component of velocity using trigonometry:
\[ v_{2y} = v_2 \sin(45^\circ) = v_2 \cdot \frac{\sqrt{2}}{2} \]
- The horizontal component \(v_{2x}\) must equal \(-10 \, \text{m/s}\), which gives us:
\[ v_{2x} = v_2 \cos(45^\circ) = v_2 \cdot \frac{\sqrt{2}}{2} = -10 \, \text{m/s} \]
- Solving for \(v_2\):
\[ v_2 = -10 \cdot \frac{\sqrt{2}}{2} = -10 \sqrt{2} \, \text{m/s} \]

5. **Finding the Time for Radius Vectors to be Perpendicular**:
- The radius vectors connecting the explosion point to each fragment will be perpendicular when their dot product is zero.
- The velocity vectors after 1 second will be:
- For the first fragment: \(v_1 = (10, 0)\)
- For the second fragment: \(v_2 = (-10\sqrt{2}/2, -10\sqrt{2}/2)\)
- The angle between two vectors can be determined using the dot product:
\[ v_1 \cdot v_2 = |v_1||v_2|\cos(\theta) \]
- Setting the dot product to zero gives us the condition for the vectors to be perpendicular.

6. **Conclusion**:
- After analyzing the motion and the angles, we find that the time taken for the two radius vectors to make a \(90^\circ\) angle is \(1 \, \text{s}\).

### Final Answer:
The time taken by the two radius vectors connecting the explosion to the fragments to make \(90^\circ\) is \(1 \, \text{s}\).