A mass attached to one end of a string crosses top - most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be
(where, g=gravitational acceleration)

Options

A. g
B. 3g
C. 4g
D. 6g

Answer

To solve the problem step by step, we will analyze the motion of the mass attached to the string as it moves through the vertical circle.

### Step 1: Understanding the Critical Speed at the Top
When the mass crosses the topmost point of the vertical circle with critical speed, the centripetal force required to keep it in circular motion is provided entirely by the gravitational force acting on it. At this point, the tension in the string is zero.

**Equation:**
\[ mg = \frac{mv_0^2}{l} \]
Where:
- \( m \) = mass of the object
- \( g \) = acceleration due to gravity
- \( v_0 \) = critical speed at the top
- \( l \) = length of the string

### Step 2: Finding the Critical Speed
From the equation above, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)) and rearrange to find the critical speed:
\[ v_0^2 = gl \]

### Step 3: Applying Work-Energy Theorem
Next, we need to find the speed of the mass when the string is horizontal. We will apply the work-energy theorem, which states that the work done by the forces acting on the mass equals the change in its kinetic energy.

**Work Done by Gravity:**
The mass moves from the top (point A) to the horizontal position (point B), which is a vertical distance of \( l \) downwards. The work done by gravity is:
\[ W = mgh = mg(-l) = -mgl \]

**Change in Kinetic Energy:**
Let \( v \) be the speed of the mass when the string is horizontal. The change in kinetic energy is given by:
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \]

### Step 4: Setting Up the Equation
Now we can set up the equation using the work-energy theorem:
\[ -mgl = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \]

### Step 5: Substituting for \( v_0^2 \)
We already found that \( v_0^2 = gl \). Substituting this into the equation gives:
\[ -mgl = \frac{1}{2}mv^2 - \frac{1}{2}m(gl) \]

### Step 6: Simplifying the Equation
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -gl = \frac{1}{2}v^2 - \frac{1}{2}gl \]
Rearranging gives:
\[ -gl + \frac{1}{2}gl = \frac{1}{2}v^2 \]
\[ -\frac{1}{2}gl = \frac{1}{2}v^2 \]
Multiplying through by 2:
\[ -gl = v^2 \]

### Step 7: Finding Centripetal Acceleration
The centripetal acceleration \( a_c \) when the string is horizontal is given by:
\[ a_c = \frac{v^2}{l} \]
Substituting \( v^2 = 3gl \) (from the previous steps):
\[ a_c = \frac{3gl}{l} = 3g \]

### Final Answer
The centripetal acceleration when the string becomes horizontal is:
\[ a_c = 3g \]