A point object moves along an arc of a circle of radius 'R'. Its velocity depends upon the distance covered 'S' as V = `Ksqrt(S)` where 'K' is a constant. If ' theta' is the angle between the total acceleration and tangential acceleration, then

Options

A. `tan theta = sqrt(S/R)`
B. `tan theta = sqrt(S/(2R))`
C. `tan theta =S/(2R)`
D. `tan theta =(2S)/R`

Answer

To solve the problem, we need to analyze the motion of the point object moving along a circular path and find the relationship between the total acceleration and the tangential acceleration.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The object moves along an arc of a circle with radius \( R \).
- The velocity \( V \) of the object is given by the equation \( V = K \sqrt{S} \), where \( K \) is a constant and \( S \) is the distance covered.

2. **Finding Tangential Acceleration**:
- Tangential acceleration \( a_t \) is defined as the rate of change of velocity with respect to time.
- We differentiate the velocity \( V \) with respect to time:
\[ a_t = \frac{dV}{dt} = \frac{d(K \sqrt{S})}{dt} \]
- Using the chain rule, we have:
\[ a_t = K \cdot \frac{1}{2} S^{-1/2} \cdot \frac{dS}{dt} = \frac{K}{2\sqrt{S}} \cdot V \]
- Since \( V = K \sqrt{S} \), substituting this in gives:
\[ a_t = \frac{K}{2\sqrt{S}} \cdot K \sqrt{S} = \frac{K^2}{2} \]

3. **Finding Normal (Centripetal) Acceleration**:
- The normal acceleration \( a_n \) (centripetal acceleration) is given by:
\[ a_n = \frac{V^2}{R} \]
- Substituting \( V = K \sqrt{S} \):
\[ a_n = \frac{(K \sqrt{S})^2}{R} = \frac{K^2 S}{R} \]

4. **Finding the Angle \( \theta \)**:
- The angle \( \theta \) between the total acceleration \( a \) and the tangential acceleration \( a_t \) can be found using the tangent of the angle:
\[ \tan(\theta) = \frac{a_n}{a_t} \]
- Substituting the expressions for \( a_n \) and \( a_t \):
\[ \tan(\theta) = \frac{\frac{K^2 S}{R}}{\frac{K^2}{2}} = \frac{2S}{R} \]

5. **Final Result**:
- Therefore, the relationship between the angle \( \theta \) and the distance \( S \) is:
\[ \tan(\theta) = \frac{2S}{R} \]

### Conclusion:
The correct answer is:
\[ \tan(\theta) = \frac{2S}{R} \]