A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as `a_(n)=kt^(2).` where k is a constant.
Tangential force on particle at t s is

Options

A. `2msqrtkR`
B. `msqrtkR`
C. `m/2sqrt(kR)`
D. `msqrt((kR)/(2))`

Answer

To solve the problem, we need to find the tangential force acting on a particle of mass \( m \) moving in a horizontal circle of radius \( R \) with a varying normal acceleration given by \( a_n = kt^2 \), where \( k \) is a constant.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The normal acceleration \( a_n \) is given as \( a_n = kt^2 \).
- The radius of the circle is \( R \).
- We need to find the tangential force acting on the particle at time \( t \).

2. **Relate Normal Acceleration to Velocity**:
- The formula for normal (centripetal) acceleration is given by:
\[ a_n = \frac{V^2}{R} \]
- Setting the two expressions for normal acceleration equal gives:
\[ kt^2 = \frac{V^2}{R} \]

3. **Solve for Velocity \( V \)**:
- Rearranging the equation for \( V^2 \):
\[ V^2 = kt^2 R \]
- Taking the square root to find \( V \):
\[ V = \sqrt{kt^2 R} = \sqrt{kR} \cdot t \]

4. **Calculate Tangential Acceleration**:
- Tangential acceleration \( a_t \) is defined as the rate of change of velocity with respect to time:
\[ a_t = \frac{dV}{dt} \]
- Since \( V = \sqrt{kR} \cdot t \), we differentiate:
\[ a_t = \frac{d}{dt}(\sqrt{kR} \cdot t) = \sqrt{kR} \cdot \frac{d}{dt}(t) = \sqrt{kR} \cdot 1 = \sqrt{kR} \]

5. **Calculate Tangential Force**:
- The tangential force \( F_t \) is given by Newton's second law:
\[ F_t = m \cdot a_t \]
- Substituting the expression for tangential acceleration:
\[ F_t = m \cdot \sqrt{kR} \]

6. **Final Answer**:
- The tangential force on the particle at time \( t \) is:
\[ F_t = m \sqrt{kR} \]