A light rod length `L`, is hanging from the vertical smooth wall of a vehicle moving with acceleration `sqrt(3)g` having a small mass attached at its one end is free to rotate about an axis passing through the other end. The minimum velocity given to the mass at its equilibrium position with respect to vehicle so that it can complete vertical circular motion respect to vehical) is `sqrt(KgL)`. Find the value of `K`.

Options

A. `sqrt(5gL)`
B. `sqrt(4gL)`
C. `sqrt(8gL)`
D. None of these

Answer

To solve the problem, we need to determine the minimum velocity that a mass attached to a light rod must have at its equilibrium position in order to complete a vertical circular motion with respect to a vehicle that is accelerating.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a light rod of length \( L \) hanging from a vertical smooth wall of a vehicle.
- The vehicle is moving with an acceleration \( a = \sqrt{3}g \).
- A small mass \( m \) is attached to one end of the rod, which can rotate about the other end (the pivot).

2. **Effective Gravity**:
- The effective acceleration acting on the mass due to the vehicle's acceleration and gravity can be calculated using vector addition.
- The gravitational acceleration \( g \) acts downwards, and the vehicle's acceleration \( \sqrt{3}g \) acts horizontally.
- The effective gravitational acceleration \( g_{\text{effective}} \) can be found using:
\[ g_{\text{effective}} = \sqrt{g^2 + (\sqrt{3}g)^2} \]
- Simplifying this gives:
\[ g_{\text{effective}} = \sqrt{g^2 + 3g^2} = \sqrt{4g^2} = 2g \]

3. **Minimum Velocity for Vertical Circular Motion**:
- For an object to complete a vertical circular motion, the minimum velocity at the top of the circle must be such that the gravitational force provides the necessary centripetal force.
- The centripetal force required at the top of the circle is given by:
\[ F_c = \frac{mv^2}{L} \]
- At the top of the circle, the gravitational force acting on the mass is \( mg_{\text{effective}} \).
- For the mass to just complete the circular motion, we set the centripetal force equal to the gravitational force:
\[ \frac{mv^2}{L} = mg_{\text{effective}} \]
- Substituting \( g_{\text{effective}} = 2g \):
\[ \frac{mv^2}{L} = m(2g) \]
- Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{L} = 2g \quad \Rightarrow \quad v^2 = 2gL \]

4. **Finding the Minimum Velocity**:
- The minimum velocity \( v \) at the bottom of the circular path (equilibrium position) can be found using energy conservation or dynamics.
- The velocity required to maintain circular motion at the bottom is given by:
\[ v_{\text{min}}^2 = 4g_{\text{effective}}L \]
- Substituting \( g_{\text{effective}} = 2g \):
\[ v_{\text{min}}^2 = 4(2g)L = 8gL \]
- Thus, the minimum velocity \( v_{\text{min}} \) is:
\[ v_{\text{min}} = \sqrt{8gL} \]

5. **Finding the Value of \( K \)**:
- We are given that the minimum velocity can be expressed as \( v_{\text{min}} = \sqrt{KgL} \).
- From our calculation, we have:
\[ \sqrt{KgL} = \sqrt{8gL} \]
- Squaring both sides gives:
\[ KgL = 8gL \]
- Dividing both sides by \( gL \) (assuming \( gL \neq 0 \)):
\[ K = 8 \]

### Final Answer:
The value of \( K \) is \( 8 \).