A smooth hemispherical bowl of radius R=0.1 m is rotating about its own axis (which is vertical) with an angular velocity 0. A particle of mass `10^(-2)` kg on the inner surface of the bowl at an angled position `theta` (with vertical) is also rotating with same '`omega`'. The particle is about a height 'h' from the bottom of thebowl.
The value of '`theta`' in temis of 'R' and h' is

Options

A. `sin^(-1)((R-h)/(R))`
B. `cos^(-1)((R-h)/(R))`
C. `tan^(-1)((R-h)/(R))`
D. `sec^(-1)((R-h)/(R))`

Answer

To solve the problem, we need to find the angle θ in terms of the radius R of the hemispherical bowl and the height h from the bottom of the bowl to the particle.

### Step-by-step Solution:

1. **Understand the Geometry**:
- We have a hemispherical bowl with radius R.
- The particle is at a height h from the bottom of the bowl.
- The angle θ is measured from the vertical axis.

2. **Identify Key Distances**:
- The distance from the center of the hemisphere to the particle (along the radius) is R.
- The height from the bottom of the bowl to the particle is h.
- Therefore, the vertical distance from the center of the hemisphere to the particle is (R - h).

3. **Construct a Right Triangle**:
- In the right triangle formed by the vertical line from the center to the particle, the horizontal distance from the center to the projection of the particle on the horizontal plane, and the line from the center to the particle:
- The vertical leg of the triangle is (R - h).
- The hypotenuse is R (the radius of the hemisphere).

4. **Use Trigonometric Relationships**:
- By the definition of cosine in a right triangle:
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{R - h}{R} \]

5. **Rearranging the Equation**:
- To find θ, we can rearrange the equation:
\[ \cos(\theta) = \frac{R - h}{R} \]
- Taking the inverse cosine gives us:
\[ \theta = \cos^{-1}\left(\frac{R - h}{R}\right) \]

### Final Result:
The angle θ in terms of R and h is:
\[ \theta = \cos^{-1}\left(\frac{R - h}{R}\right) \]