A uniform rope of mass `0.1` kg and length `2.5` m
hangs from ceiling. The speed of transverse wave in
the rope at upper end at a point `0.5` m distance
from lower end will be

Options

A. `5" ms"^(-1) and 2.24 "ms"^(-1)`
B. `10" ms"^(-1) and 3.23 "ms"^(-1)`
C. `7.5" ms"^(-1) and 1.2 "ms"^(-1)`
D. None of these

Explanation

To solve the problem of finding the speed of transverse waves in a hanging rope at a specific point, we will follow these steps:

### Step 1: Understand the Problem
We have a uniform rope of mass \( m = 0.1 \) kg and length \( L = 2.5 \) m hanging from the ceiling. We need to find the speed of transverse waves at a point \( 0.5 \) m from the lower end of the rope.

### Step 2: Calculate the Linear Mass Density
The linear mass density \( \mu \) of the rope can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{0.1 \, \text{kg}}{2.5 \, \text{m}} = 0.04 \, \text{kg/m} \]

### Step 3: Determine the Tension in the Rope
The tension \( T \) at a distance \( x \) from the lower end of the rope is due to the weight of the rope below that point. The length of the rope below the point \( 0.5 \) m from the lower end is:
\[ L - x = 2.5 \, \text{m} - 0.5 \, \text{m} = 2.0 \, \text{m} \]
The mass of the rope below this point is:
\[ m_{\text{below}} = \mu \cdot (L - x) = 0.04 \, \text{kg/m} \cdot 2.0 \, \text{m} = 0.08 \, \text{kg} \]
The tension \( T \) at point \( 0.5 \) m from the lower end is given by:
\[ T = m_{\text{below}} \cdot g = 0.08 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.8 \, \text{N} \]

### Step 4: Calculate the Speed of the Wave
The speed \( v \) of a transverse wave in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values we calculated:
\[ v = \sqrt{\frac{0.8 \, \text{N}}{0.04 \, \text{kg/m}}} = \sqrt{20} \approx 4.47 \, \text{m/s} \]

### Final Answer
The speed of the transverse wave at a point \( 0.5 \) m from the lower end of the rope is approximately \( 4.47 \, \text{m/s} \).

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