A disc of mass 0.5kg is kept floating horizontally in mid air by firing bullets of mass 5 g each, vertically at it at the rate of 10 per second. If the bullets drop dead. The speed of the bullet striking the disc is (g=10 `ms^(-2)`)

Options

A. `100 ms^(-1)`
B. `10 ms^(-1)`
C. `10 cm//s`
D. `1 cm//s`

Answer

To solve the problem, we need to find the speed of the bullets striking the disc. We will use the principles of momentum and force.

### Step-by-Step Solution:

1. **Identify the mass of the disc and the bullets**:
- Mass of the disc, \( M = 0.5 \, \text{kg} \)
- Mass of each bullet, \( m = 5 \, \text{g} = 0.005 \, \text{kg} \)

2. **Determine the rate of bullets fired**:
- Bullets fired per second = 10

3. **Calculate the total mass of bullets hitting the disc per second**:
- Total mass of bullets hitting the disc per second = \( 10 \times m = 10 \times 0.005 \, \text{kg} = 0.05 \, \text{kg} \)

4. **Calculate the weight of the disc**:
- Weight of the disc, \( W = M \times g = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \)

5. **Use the principle of momentum**:
- The change in momentum per second due to the bullets can be expressed as:
\[ \text{Change in momentum} = \text{Total mass of bullets per second} \times \text{velocity of bullets} = 0.05 \, \text{kg} \times v \]

6. **Set the upward force equal to the weight of the disc**:
- The force exerted by the bullets must equal the weight of the disc for it to remain floating:
\[ \text{Force} = \text{Change in momentum per second} = 5 \, \text{N} \]
- Therefore, we have:
\[ 0.05 \, v = 5 \]

7. **Solve for \( v \)**:
- Rearranging the equation gives:
\[ v = \frac{5}{0.05} = 100 \, \text{m/s} \]

### Final Answer:
The speed of the bullet striking the disc is \( 100 \, \text{m/s} \).