A thin rod of length `1 m` is fixed in a vertical position inside a train, which is moving horizentally with constant acceleration `4 ms^(-2)`.A bead can slide on the rod and friction coefficient between them is `0.5`. If the head is released from rest at the top of the rod , it will reach the bottom the in

Options

A. `sqrt2 s`
B. `1 s`
C. `2 s`
D. `0.5 s`

Explanation

To solve the problem, we need to analyze the forces acting on the bead as it slides down the rod in the accelerating train. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the forces acting on the bead
The bead experiences the following forces:
- Gravitational force (\( mg \)) acting downward.
- Normal force (\( N \)) acting perpendicular to the rod.
- Frictional force (\( F_f \)) acting along the rod, opposing the motion of the bead.

### Step 2: Determine the pseudo force
Since the train is accelerating, a pseudo force acts on the bead in the opposite direction of the train's acceleration. The magnitude of this pseudo force is:
\[ F_{pseudo} = m \cdot a \]
where \( a = 4 \, \text{m/s}^2 \) (the acceleration of the train).

### Step 3: Calculate the normal force
The normal force \( N \) can be equated to the pseudo force acting on the bead:
\[ N = m \cdot a = m \cdot 4 \]

### Step 4: Calculate the frictional force
The frictional force \( F_f \) is given by:
\[ F_f = \mu \cdot N = \mu \cdot (m \cdot 4) \]
where \( \mu = 0.5 \) (the coefficient of friction). Thus,
\[ F_f = 0.5 \cdot (m \cdot 4) = 2m \]

### Step 5: Determine the net vertical acceleration
The net vertical acceleration \( A_v \) of the bead can be expressed as:
\[ A_v = g - \frac{F_f}{m} \]
Substituting the values:
\[ A_v = g - 2 = 10 - 2 = 8 \, \text{m/s}^2 \]

### Step 6: Use kinematic equations to find the time taken to slide down the rod
The bead starts from rest and slides down a distance \( s = 1 \, \text{m} \) with an acceleration \( A_v = 8 \, \text{m/s}^2 \). Using the kinematic equation:
\[ s = ut + \frac{1}{2} A_v t^2 \]
Since the initial velocity \( u = 0 \):
\[ 1 = 0 + \frac{1}{2} \cdot 8 \cdot t^2 \]
This simplifies to:
\[ 1 = 4t^2 \]
\[ t^2 = \frac{1}{4} \]
\[ t = \frac{1}{2} \, \text{s} \]

### Conclusion
The time taken for the bead to reach the bottom of the rod is \( 0.5 \, \text{s} \).