A pendulum consists of a wooden bob of mass `M` and length `l`. A bullet of mass `m` is fired towards the pendulum with a speed `v`. The bullet emerges immediately out of the bob from the other side with a speed of `v//2` and the bob starts rising. Assume no loss of mass of bob takes place due to penetration.
If the bob is just able to complete the circular motion, then tension at the lowest point just when the bob starts rising will be

Options

A. `6Mg`
B. `5Mg`
C. `3Mg`
D. `Mg`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the scenario
A pendulum consists of a wooden bob of mass \( M \) and length \( l \). A bullet of mass \( m \) is fired towards the pendulum with a speed \( v \) and emerges from the bob with a speed of \( \frac{v}{2} \). We need to find the tension in the string at the lowest point of the pendulum's swing when the bob starts rising.

### Step 2: Analyze the motion at the top of the swing
For the bob to just complete the circular motion, at the highest point of the swing, the tension \( T \) in the string will be zero. The forces acting on the bob at this point are:
- Weight of the bob: \( Mg \)
- Centripetal force required to keep the bob in circular motion: \( \frac{Mv_1^2}{l} \)

Setting the centripetal force equal to the weight:
\[ \frac{Mv_1^2}{l} = Mg \]
This simplifies to:
\[ v_1^2 = gl \]

### Step 3: Calculate the velocity of the bob just after the bullet passes through it
Using the principle of conservation of momentum, the momentum before the bullet hits the bob is equal to the momentum after the bullet exits:
\[ mv = (M + m)v_b + mv_b' \]
where \( v_b \) is the velocity of the bob after the bullet enters, and \( v_b' = \frac{v}{2} \) is the velocity of the bullet after it exits.

Rearranging gives:
\[ mv = Mv_b + mv_b' \]
Substituting \( v_b' \):
\[ mv = Mv_b + m\left(\frac{v}{2}\right) \]
\[ mv - \frac{mv}{2} = Mv_b \]
\[ \frac{mv}{2} = Mv_b \implies v_b = \frac{mv}{2M} \]

### Step 4: Calculate the velocity of the bob at the lowest point
Using conservation of energy between the lowest point and the point just after the bullet exits:
\[ \frac{1}{2}Mv_b^2 + mgh = \frac{1}{2}M(0)^2 + mgh \]
At the lowest point, the total energy is:
\[ \frac{1}{2}M(0)^2 + mgh + \frac{1}{2}mv_b'^2 \]

### Step 5: Calculate the tension at the lowest point
At the lowest point, the forces acting on the bob are:
- Tension \( T \) upwards
- Weight \( Mg \) downwards
- Centripetal force \( \frac{Mv^2}{l} \)

Using the equation of motion:
\[ T - Mg = \frac{Mv^2}{l} \]
Thus,
\[ T = Mg + \frac{Mv^2}{l} \]

### Step 6: Substitute the values
We already found that \( v_b = \frac{mv}{2M} \) and \( v_1^2 = gl \).
Substituting \( v^2 \) into the tension equation gives:
\[ T = Mg + \frac{M(5gL)}{l} \]

### Final Step: Conclusion
The tension at the lowest point when the bob starts rising is:
\[ T = 6Mg \]