A uniform rope of length `20`m and mass `5`kg is hanging vertically from a rigid support. A block of mass `4`kg is attached to the free end. The wave length of the transverse wave pulse at the lower end of the rope is `0.04` m. The wavelength of the same pulse as it reaches the top is

Options

A. `0.06m`
B. `0.12m`
C. `1.5m`
D. `2.2m`

Explanation

To solve the problem, we need to find the wavelength of the transverse wave pulse as it reaches the top of the rope. We will follow these steps:

### Step 1: Understand the Setup
We have a rope of length 20 m and mass 5 kg hanging vertically with a block of mass 4 kg attached to the free end. The tension in the rope will vary along its length due to the weight of the rope and the block.

### Step 2: Calculate the Tension at the Bottom of the Rope
At the bottom of the rope, the tension (T1) is equal to the weight of the block:
\[ T_1 = m \cdot g = 4 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 3: Calculate the Velocity of the Wave at the Bottom of the Rope
The velocity (V1) of the wave at the bottom can be calculated using the formula:
\[ V_1 = \sqrt{\frac{T_1}{\mu}} \]
where \(\mu\) (mass per unit length) is given by:
\[ \mu = \frac{\text{mass of rope}}{\text{length of rope}} = \frac{5 \, \text{kg}}{20 \, \text{m}} = 0.25 \, \text{kg/m} \]

Now substituting the values:
\[ V_1 = \sqrt{\frac{40 \, \text{N}}{0.25 \, \text{kg/m}}} = \sqrt{160} = 4\sqrt{10} \, \text{m/s} \]

### Step 4: Calculate the Tension at the Top of the Rope
At the top of the rope, the tension (T2) is equal to the weight of the block plus the weight of the rope:
\[ T_2 = (m + \text{mass of rope}) \cdot g = (4 \, \text{kg} + 5 \, \text{kg}) \cdot 10 \, \text{m/s}^2 = 90 \, \text{N} \]

### Step 5: Calculate the Velocity of the Wave at the Top of the Rope
Using the same formula for velocity:
\[ V_2 = \sqrt{\frac{T_2}{\mu}} \]
Substituting the values:
\[ V_2 = \sqrt{\frac{90 \, \text{N}}{0.25 \, \text{kg/m}}} = \sqrt{360} = 6\sqrt{10} \, \text{m/s} \]

### Step 6: Relate Wavelengths and Frequencies
The frequency of the wave remains constant as it travels through the rope. Thus, we can write:
\[ V_1 = \lambda_1 \cdot f \]
\[ V_2 = \lambda_2 \cdot f \]

### Step 7: Find the Relationship Between Wavelengths
Dividing the two equations:
\[ \frac{V_2}{V_1} = \frac{\lambda_2}{\lambda_1} \]
Substituting the values of velocities:
\[ \frac{6\sqrt{10}}{4\sqrt{10}} = \frac{\lambda_2}{0.04} \]

### Step 8: Solve for \(\lambda_2\)
The \(\sqrt{10}\) cancels out:
\[ \frac{6}{4} = \frac{\lambda_2}{0.04} \]
\[ \lambda_2 = \frac{6}{4} \cdot 0.04 = 0.06 \, \text{m} \]

### Final Answer
The wavelength of the same pulse as it reaches the top of the rope is \( \lambda_2 = 0.06 \, \text{m} \).

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