A particle of mass m is revolving in a horizontal circle on a smooth table by an in extensible massless string if the initial speed of the particle is `V_(0)` and the particle has tangential acceleration of constant magnitude a,
If the breaking strength of the string is equal to weight of the stone. The time after which the string will break

Options

A. `(1)/(a)(sqrt(gR)-V_(0))`
B. `(V_(0))/(a)(sqrt(gR)-V_(0))`
C. `(1)/(a)(2gR-V_(0))`
D. `(V_(0))/(a)(2gR-V_(0))`

Answer

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the Forces Acting on the Particle
The particle of mass \( m \) is revolving in a horizontal circle. The forces acting on it include:
- Tension \( T \) in the string, which provides the centripetal force.
- Weight \( mg \) acting downwards.

### Step 2: Write the Expression for Tension
The tension in the string must provide the necessary centripetal force to keep the particle moving in a circle. The centripetal force is given by:
\[ T = \frac{mv^2}{r} \]
where \( v \) is the speed of the particle and \( r \) is the radius of the circular path.

### Step 3: Condition for the String to Break
The string will break when the tension equals the weight of the particle:
\[ T = mg \]
Setting the two expressions for tension equal gives:
\[ \frac{mv^2}{r} = mg \]

### Step 4: Simplify the Equation
We can simplify the equation by canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r} = g \]
This implies:
\[ v^2 = gr \]

### Step 5: Relate Speed to Time
The particle has an initial speed \( V_0 \) and a constant tangential acceleration \( a \). The speed of the particle at any time \( t \) can be expressed as:
\[ v = V_0 + at \]

### Step 6: Set the Speed Equal to the Breaking Condition
At the time the string breaks, we set the speed equal to the speed derived from the tension condition:
\[ V_0 + at = \sqrt{gr} \]

### Step 7: Solve for Time \( t \)
Rearranging the equation to solve for \( t \):
\[ at = \sqrt{gr} - V_0 \]
\[ t = \frac{\sqrt{gr} - V_0}{a} \]

### Final Result
The time after which the string will break is:
\[ t = \frac{\sqrt{gr} - V_0}{a} \]