A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is held in contact with the inner side of upper rim at a point P. It is given a horizontal speed `v_(0)` tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be [Take acceleration due to gravity g]

Options

A. `(h)/(2piR)`
B. `(v_(0))/(h)sqrt((h)/(2g))`
C. `(v_(0))/(2piR)sqrt((2h)/(g))`
D. `(v_(0)pi)/(g)sqrt((2gh)/(2R))`

Answer

To solve the problem of determining the number of revolutions made by a particle inside a hollow vertical cylinder, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a hollow vertical cylinder with radius \( R \) and height \( h \).
- A particle is placed at the top rim (point P) and given a horizontal speed \( v_0 \) tangential to the rim.

2. **Motion of the Particle**:
- As the particle moves, it will experience two types of motion: horizontal (tangential) motion and vertical motion due to gravity.
- The particle will leave the cylinder at point Q, which is directly below point P.

3. **Vertical Motion Analysis**:
- The vertical motion of the particle can be described using the equation of motion under gravity:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging this gives us the time \( t \) taken to fall a height \( h \):
\[ t = \sqrt{\frac{2h}{g}} \]

4. **Horizontal Motion Analysis**:
- The horizontal distance traveled by the particle while it falls is determined by its initial horizontal speed \( v_0 \):
\[ \text{Horizontal distance} = v_0 \cdot t \]
- Substituting the expression for \( t \):
\[ \text{Horizontal distance} = v_0 \cdot \sqrt{\frac{2h}{g}} \]

5. **Circumference of the Cylinder**:
- The circumference of the cylinder (the distance covered in one complete revolution) is given by:
\[ C = 2 \pi R \]

6. **Number of Revolutions**:
- The number of revolutions \( n \) made by the particle can be calculated by dividing the horizontal distance by the circumference:
\[ n = \frac{\text{Horizontal distance}}{C} = \frac{v_0 \cdot \sqrt{\frac{2h}{g}}}{2 \pi R} \]
- Simplifying this gives:
\[ n = \frac{v_0}{2 \pi R} \cdot \sqrt{\frac{2h}{g}} \]

### Final Result:
The number of revolutions made by the particle is:
\[ n = \frac{v_0}{2 \pi R} \cdot \sqrt{\frac{2h}{g}} \]