One end of a string of length L is tied to the celling of lift accelerating upwards with an accelerating 2g The other end of the string is free. The linear mass density of the string varies linearly from 0 of `lamda` from bottom to top :-

Options

A. The velocity of the wave in the string will be 0
B. The acceleration of the wave in the string will be `3g//4` every where
C. The time taken by a pulse to reach from bottom to top will be `sqrt(8 L//3g)`
D. The time taken by a pulse to reach from bottom to top will be `sqrt(4 L//3g)`

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understanding the Problem
We have a string of length \( L \) tied to the ceiling of a lift that is accelerating upwards with an acceleration of \( 2g \). The linear mass density of the string varies linearly from \( 0 \) at the top to \( \lambda \) at the bottom. We need to find the velocity of the wave in the string.

### Step 2: Determine the Linear Mass Density
The linear mass density \( \mu(y) \) at a distance \( y \) from the bottom of the string can be expressed as:
\[ \mu(y) = \frac{\lambda}{L} y \]
This is because the mass density varies linearly from \( 0 \) to \( \lambda \) as we move from the top to the bottom of the string.

### Step 3: Calculate the Tension in the String
To find the tension \( T \) in the string, we consider a small element of the string of length \( dy \) at a distance \( y \) from the bottom. The mass of this small element is:
\[ dm = \mu(y) \cdot dy = \frac{\lambda}{L} y \cdot dy \]
The forces acting on this element are the tension \( T \) acting upwards and the weight of the element acting downwards, which is \( dm \cdot g \). Since the lift is accelerating upwards with \( 2g \), we can set up the equation:
\[ T - dm \cdot g = dm \cdot (2g) \]
Substituting \( dm \):
\[ T - \left(\frac{\lambda}{L} y \cdot dy\right) g = \left(\frac{\lambda}{L} y \cdot dy\right) (2g) \]
Rearranging gives:
\[ T = 3 \left(\frac{\lambda}{L} y \cdot dy\right) g \]

### Step 4: Substitute for the Tension
Now we can express the tension \( T \) in terms of \( y \):
\[ T = \frac{3\lambda g y}{L} \]

### Step 5: Calculate the Velocity of the Wave
The velocity \( v \) of the wave in the string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the expressions for \( T \) and \( \mu \):
\[ v = \sqrt{\frac{\frac{3\lambda g y}{L}}{\frac{\lambda}{L} y}} = \sqrt{\frac{3g y}{1}} = \sqrt{3gy} \]

### Step 6: Find the Acceleration of the Wave
The acceleration \( a \) of the wave can be found by differentiating the velocity with respect to \( y \):
\[ a = \frac{dv}{dt} = v \frac{dv}{dy} \]
Calculating \( \frac{dv}{dy} \):
\[ \frac{dv}{dy} = \frac{3g}{2\sqrt{3gy}} \]
Thus, the acceleration becomes:
\[ a = \sqrt{3gy} \cdot \frac{3g}{2\sqrt{3gy}} = \frac{3g}{2} \]

### Step 7: Time Taken for a Pulse to Travel from Bottom to Top
To find the time \( t \) taken for a pulse to travel from the bottom to the top of the string, we can use:
\[ v = \frac{dy}{dt} \]
Rearranging gives:
\[ dt = \frac{dy}{\sqrt{3gy}} \]
Integrating from \( 0 \) to \( L \):
\[ t = \int_0^L \frac{dy}{\sqrt{3gy}} = \frac{2}{3\sqrt{g}} \left[ \sqrt{y} \right]_0^L = \frac{2}{3\sqrt{g}} \sqrt{L} \]
Thus, the total time taken is:
\[ t = \frac{8L}{3g} \]

### Final Answers
1. The velocity of the wave in the string is \( \sqrt{3gy} \).
2. The time taken for a pulse to travel from the bottom to the top of the string is \( \frac{8L}{3g} \).