A uniform rope of mass 6 kg and length 6 m hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.075 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is K cm where K is

Answer

To solve the problem, we need to find the wavelength of a transverse pulse when it reaches the top of a uniform rope that is hanging vertically with a block attached at the end. Here's a step-by-step solution:

### Step 1: Understand the System
We have a uniform rope of mass \( m = 6 \, \text{kg} \) and length \( L = 6 \, \text{m} \) hanging vertically. A block of mass \( M = 2 \, \text{kg} \) is attached to the free end of the rope.

### Step 2: Calculate the Mass per Unit Length (\( \mu \))
The mass per unit length (\( \mu \)) of the rope can be calculated as:
\[ \mu = \frac{m}{L} = \frac{6 \, \text{kg}}{6 \, \text{m}} = 1 \, \text{kg/m} \]

### Step 3: Determine the Tension in the Rope
The tension in the rope varies along its length due to the weight of the rope and the block. The tension at a distance \( x \) from the bottom of the rope can be expressed as:
\[ T(x) = M \cdot g + \text{Weight of the rope below } x \]
Where the weight of the rope below point \( x \) is given by:
\[ \text{Weight} = \mu \cdot x \cdot g \]
Thus, the total tension at point \( x \) is:
\[ T(x) = 2g + \mu \cdot x \cdot g = 2g + 1 \cdot x \cdot g = (2 + x)g \]

### Step 4: Calculate the Wave Velocity
The velocity of a transverse wave in the rope is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting \( T(x) \):
\[ v(x) = \sqrt{\frac{(2 + x)g}{1}} = \sqrt{(2 + x)g} \]

### Step 5: Relate Wavelength and Velocity
The relationship between velocity, frequency, and wavelength is given by:
\[ v = f \lambda \]
Since the frequency \( f \) remains constant, we can write:
\[ \frac{\lambda_H}{\lambda_L} = \frac{v_H}{v_L} \]
Where:
- \( \lambda_H \) is the wavelength at the top of the rope,
- \( \lambda_L = 0.075 \, \text{m} \) is the wavelength at the bottom of the rope,
- \( v_H \) is the velocity at the top of the rope,
- \( v_L \) is the velocity at the bottom of the rope.

### Step 6: Calculate Velocities at the Ends
1. **At the bottom of the rope** (\( x = 0 \)):
\[ v_L = \sqrt{(2 + 0)g} = \sqrt{2g} \]

2. **At the top of the rope** (\( x = 6 \)):
\[ v_H = \sqrt{(2 + 6)g} = \sqrt{8g} \]

### Step 7: Substitute and Solve for \( \lambda_H \)
Now substituting into the wavelength relationship:
\[ \frac{\lambda_H}{0.075} = \frac{\sqrt{8g}}{\sqrt{2g}} \]
This simplifies to:
\[ \frac{\lambda_H}{0.075} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2 \]
Thus,
\[ \lambda_H = 2 \times 0.075 = 0.150 \, \text{m} \]

### Step 8: Convert to Centimeters
To convert meters to centimeters:
\[ \lambda_H = 0.150 \, \text{m} \times 100 = 15 \, \text{cm} \]

### Final Answer
The value of \( K \) is:
\[ K = 15 \]