A uniform rope of length 12 mm and mass 6 kg 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 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

Options

A. 12
B. 3
C. 9
D. 6

Explanation

To solve the problem, we need to determine the wavelength of a transverse pulse as it travels from the bottom to the top of a hanging rope. The key steps involve calculating the tension at different points in the rope and using the relationship between wave speed, tension, and wavelength.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the rope (L) = 12 mm = 0.012 m
- Mass of the rope (m_rope) = 6 kg
- Mass of the block (m_block) = 2 kg
- Wavelength at the bottom of the rope (λ_A) = 0.06 m

2. **Calculate the Tension at the Bottom of the Rope (T_A):**
- The tension at the bottom of the rope is equal to the weight of the block.
- \( T_A = m_{block} \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \)

3. **Calculate the Total Weight at the Top of the Rope (T_B):**
- The total weight acting on the rope includes the weight of the block and the weight of the rope itself.
- The weight of the rope is \( m_{rope} \cdot g = 6 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 58.8 \, \text{N} \)
- Therefore, the total tension at the top of the rope is:
- \( T_B = T_A + m_{rope} \cdot g = 19.6 \, \text{N} + 58.8 \, \text{N} = 78.4 \, \text{N} \)

4. **Determine the Wave Speeds at Both Points:**
- The speed of a wave on a string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( \mu \) is the mass per unit length.
- The mass per unit length of the rope is \( \mu = \frac{m_{rope}}{L} = \frac{6 \, \text{kg}}{0.012 \, \text{m}} = 500 \, \text{kg/m} \)

5. **Calculate the Wave Speeds:**
- Speed at the bottom (v_A):
\[ v_A = \sqrt{\frac{T_A}{\mu}} = \sqrt{\frac{19.6 \, \text{N}}{500 \, \text{kg/m}}} = \sqrt{0.0392} \approx 0.198 \, \text{m/s} \]
- Speed at the top (v_B):
\[ v_B = \sqrt{\frac{T_B}{\mu}} = \sqrt{\frac{78.4 \, \text{N}}{500 \, \text{kg/m}}} = \sqrt{0.1568} \approx 0.396 \, \text{m/s} \]

6. **Use the Relationship Between Wavelength and Wave Speed:**
- Since the frequency (f) remains constant, we can use the relationship \( v = f \cdot \lambda \).
- The ratio of the wavelengths is given by:
\[ \frac{\lambda_A}{\lambda_B} = \frac{v_A}{v_B} \]
- Substituting the values:
\[ \frac{0.06 \, \text{m}}{\lambda_B} = \frac{0.198 \, \text{m/s}}{0.396 \, \text{m/s}} \implies \lambda_B = 2 \cdot 0.06 \, \text{m} = 0.12 \, \text{m} \]

### Final Answer:
The wavelength of the pulse when it reaches the top of the rope is **0.12 m**.