A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tension at the bottom and top would be-

Options

A. 6 mg
B. 4 mg
C. 3 mg
D. 2 mg

Explanation

To solve the problem of finding the difference in tension at the bottom and top of a particle rotating in a vertical circle, we can follow these steps:

### Step 1: Identify Forces at the Top and Bottom
At the topmost point of the vertical circle, the forces acting on the particle are:
- Weight (mg) acting downward
- Tension (T1) in the string acting downward

At the bottommost point, the forces are:
- Weight (mg) acting downward
- Tension (T2) in the string acting upward

### Step 2: Apply Newton's Second Law at the Top
At the topmost point, the net force acting on the particle provides the centripetal force required for circular motion. Therefore, we can write:
\[ T1 + mg = \frac{mv^2}{l} \]
Where:
- \( v \) is the velocity at the top
- \( l \) is the length of the string

Rearranging gives us:
\[ T1 = \frac{mv^2}{l} - mg \]

### Step 3: Apply Newton's Second Law at the Bottom
At the bottommost point, the net force also provides the centripetal force:
\[ T2 - mg = \frac{mu^2}{l} \]
Where:
- \( u \) is the velocity at the bottom

Rearranging gives us:
\[ T2 = \frac{mu^2}{l} + mg \]

### Step 4: Find the Relationship Between Velocities
To relate \( v \) and \( u \), we can use the conservation of energy or the work-energy theorem. The change in kinetic energy as the particle moves from the bottom to the top is equal to the work done against gravity:
\[ \Delta KE = Work_{gravity} \]
The work done by gravity when moving a distance of \( 2l \) (upward) is:
\[ Work_{gravity} = -mg(2l) = -2mgl \]
Thus, we have:
\[ \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = -2mgl \]
This simplifies to:
\[ v^2 - u^2 = -4gl \]

### Step 5: Substitute the Velocity Relationship into Tension Equations
Now, substituting \( v^2 - u^2 = -4gl \) into the equations for \( T1 \) and \( T2 \):
1. For \( T1 \):
\[ T1 = \frac{m(-4gl + u^2)}{l} - mg \]
\[ T1 = \frac{mu^2}{l} - \frac{4mg}{l} - mg \]
\[ T1 = \frac{mu^2}{l} - 5mg \]

2. For \( T2 \):
\[ T2 = \frac{mu^2}{l} + mg \]

### Step 6: Calculate the Difference in Tension
Now, we can find the difference \( T2 - T1 \):
\[ T2 - T1 = \left( \frac{mu^2}{l} + mg \right) - \left( \frac{mu^2}{l} - 5mg \right) \]
This simplifies to:
\[ T2 - T1 = mg + 5mg = 6mg \]

### Final Answer
Thus, the difference in tension at the bottom and top is:
\[ T2 - T1 = 6mg \]