A dics is rotating in a room. A boy standing near the rim of the disc of radius `R` finds the water droplet falling from the ceiling is always falling on his head. As one drop hits his head, other one starts from the ceiling. If height of the roof above his head is `H` , then angular velocity of the disc is

Options

A. `pisqrt((2gR)/(H^(2))`
B. `pisqrt((2gH)/(R^(2))`
C. `pisqrt((2g)/(H))`
D. None of these

Explanation

To solve the problem, we need to determine the angular velocity (\( \omega \)) of a disc that is rotating such that water droplets falling from the ceiling always land on the boy's head standing at the rim of the disc.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The boy is standing at the rim of a disc of radius \( R \).
- Water droplets fall from the ceiling (height \( H \)) and hit the boy's head.
- The time taken for a droplet to fall from the ceiling to the boy's head is denoted as \( T \).

2. **Time of Fall**:
- The time \( T \) for the droplet to fall can be calculated using the equation of motion under gravity:
\[ H = \frac{1}{2} g T^2 \]
- Rearranging this gives:
\[ T^2 = \frac{2H}{g} \]
- Taking the square root:
\[ T = \sqrt{\frac{2H}{g}} \]

3. **Relating Time to Angular Velocity**:
- The disc makes one complete rotation in time \( T \). The angular velocity \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the expression for \( T \):
\[ \omega = \frac{2\pi}{\sqrt{\frac{2H}{g}}} \]

4. **Simplifying the Expression**:
- We can simplify this expression further:
\[ \omega = 2\pi \cdot \sqrt{\frac{g}{2H}} \]
- This can be rewritten as:
\[ \omega = \pi \cdot \sqrt{\frac{4g}{2H}} = \pi \cdot \sqrt{\frac{2g}{H}} \]

5. **Final Result**:
- Thus, the angular velocity of the disc is:
\[ \omega = \pi \sqrt{\frac{2g}{H}} \]