To solve the problem step by step, we will follow the logical flow of the video transcript while providing clear explanations and calculations.
### Step 1: Understanding the System
We have a stainless steel wire AB of length 10 m and mass 1 kg, with one end fixed at point A. The other end is connected to a massless string BC that goes over a smooth pulley and is attached to a container of mass 2 kg. Water is poured into the container at a rate of 2.25 liters per second.
### Step 2: Finding the Mass per Unit Length (μ) of the Wire
The mass per unit length (μ) of the wire can be calculated as follows:
\[
\mu = \frac{\text{mass of wire}}{\text{length of wire}} = \frac{1 \, \text{kg}}{10 \, \text{m}} = 0.1 \, \text{kg/m}
\]
### Step 3: Finding the Rate of Mass Increase in the Container
The rate at which water is being poured into the container is given as 2.25 liters per second. We convert this to cubic meters:
\[
2.25 \, \text{liters/sec} = 2.25 \times 10^{-3} \, \text{m}^3/\text{sec}
\]
The mass of water added per second can be calculated using the density of water (1000 kg/m³):
\[
\text{mass rate} = \text{density} \times \text{volume rate} = 1000 \, \text{kg/m}^3 \times 2.25 \times 10^{-3} \, \text{m}^3/\text{sec} = 2.25 \, \text{kg/sec}
\]
### Step 4: Expression for Tension (T) in the String
At any time \( t \), the total mass in the container is the sum of the mass of the container and the mass of the water:
\[
\text{Total mass} = 2 \, \text{kg} + 2.25 \, \text{kg/sec} \times t
\]
The tension in the string due to the weight of this mass is:
\[
T(t) = g \times \text{Total mass} = 10 \, \text{m/s}^2 \times (2 + 2.25t) = 20 + 22.5t \, \text{N}
\]
### Step 5: Finding the Velocity (v) of the Wave in the Wire
The velocity of the wave in the wire is given by the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
Substituting the expression for tension:
\[
v(t) = \sqrt{\frac{20 + 22.5t}{0.1}} = \sqrt{200 + 225t}
\]
### Step 6: Time Taken for the Wave to Travel 10 m
We need to find the time \( T \) taken for the wave to travel from A to B (10 m). We can express the velocity as:
\[
\frac{dx}{dt} = \sqrt{200 + 225t}
\]
Integrating both sides, we have:
\[
\int_0^{10} dx = \int_0^{T} \sqrt{200 + 225t} \, dt
\]
The left side integrates to 10, and the right side can be solved using substitution.
### Step 7: Solving the Integral
Let \( u = 200 + 225t \), then \( du = 225 \, dt \) or \( dt = \frac{du}{225} \).
When \( t = 0 \), \( u = 200 \), and when \( t = T \), \( u = 200 + 225T \).
The integral becomes:
\[
10 = \int_{200}^{200 + 225T} \sqrt{u} \frac{du}{225}
\]
Calculating this integral gives:
\[
10 = \frac{1}{225} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{200}^{200 + 225T}
\]
Solving for \( T \) will yield the time taken for the wave to reach point B.
### Step 8: Finding the Tension at Time T
Once we have \( T \), we can substitute it back into the tension equation:
\[
T = 20 + 22.5T
\]
This will give us the tension in the string when the wave reaches point B.
### Final Results
After performing the calculations, we find:
- The time taken for the wave to reach point B is approximately \( 0.612 \, \text{seconds} \).
- The tension in the string at that time is approximately \( 33.77 \, \text{N} \).
