A 10 m long horizontal stainless steel wire AB of mass 1 kg whose end A is fixed, is connected to a massless string BC passing over a smooth pulley. String BC is connected to a container of mass 2 kg at end C. Water (density `=1xx10^(3)kg//m^(3)`) is poured int he container at a constant rate of 2.25 litre/sec at t=0. also t=0, a puls is generated at end A.

Options

A. Time taken by the pulse to reach point B is 0.612 sec
B. Time taken by the pulse to reach point B is 0.212 sec
C. Tension in the string at this moment is 33.77 N
D. Tension in the string at this moment is 24.77 N

Explanation

To solve the problem step by step, we will analyze the situation involving the stainless steel wire, the mass, and the water being poured into the container. We will determine the time taken for the pulse generated at point A to reach point B.

### Step 1: Understanding the Setup
We have:
- A stainless steel wire AB of length 10 m and mass 1 kg, fixed at point A.
- A massless string BC over a smooth pulley, connected to a container of mass 2 kg at point C.
- Water is poured into the container at a rate of 2.25 liters/sec (which is equivalent to 2.25 kg/sec since the density of water is 1000 kg/m³).
- A pulse is generated at point A at time t = 0.

### Step 2: Calculate the Mass per Unit Length (μ) of the Wire
The mass per unit length (μ) of the wire is calculated as:
\[ \mu = \frac{\text{mass of wire}}{\text{length of wire}} = \frac{1 \text{ kg}}{10 \text{ m}} = 0.1 \text{ kg/m} \]

### Step 3: Determine the Tension in the String
The tension in the string will change as water is added to the container. The weight of the container increases with the mass of water added. The mass of water at time \( t \) is given by:
\[ \text{mass of water} = 2.25 \text{ kg/s} \times t \]
Thus, the total mass in the container at time \( t \) is:
\[ \text{Total mass} = 2 \text{ kg} + 2.25 \text{ kg/s} \times t \]
The tension \( T(t) \) in the string is equal to the weight of the container:
\[ T(t) = \left(2 + 2.25t\right) \times g \]
Assuming \( g = 10 \text{ m/s}^2 \):
\[ T(t) = (2 + 2.25t) \times 10 = 20 + 22.5t \text{ N} \]

### Step 4: Calculate the Speed of the Pulse in the Wire
The speed \( v \) of the wave in the wire is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting \( T(t) \) and \( \mu \):
\[ v(t) = \sqrt{\frac{20 + 22.5t}{0.1}} = \sqrt{200 + 225t} \]

### Step 5: Set Up the Integral to Find Time
To find the time \( t \) taken for the pulse to travel from A to B (10 m), we use the relationship:
\[ dx = v(t) dt \]
Integrating from \( x = 0 \) to \( x = 10 \) and \( t = 0 \) to \( t \):
\[ \int_0^{10} dx = \int_0^t v(t) dt \]
This gives:
\[ 10 = \int_0^t \sqrt{200 + 225t} dt \]

### Step 6: Solve the Integral
To solve the integral, we can use substitution or numerical methods. However, for simplicity, we can evaluate it directly using numerical integration techniques or computational tools.

After performing the integration, we find that:
\[ t \approx 0.612 \text{ seconds} \]

### Step 7: Conclusion
The time taken by the pulse to reach point B from point A is approximately **0.612 seconds**.