identical wires A and B of different materials are hung from then ceiling of a room . The density of wire A is greater than the density of wire B but their lengths are same. Identical wave pulses are produced at the bottom of respective wires. The time taken by the pulse to reach the top is

Options

A. greater for wire A
B. greater for wire B
C. same for both the wires
D. cannot be determind

Answer

To solve the problem, we need to analyze the wave motion in the two wires A and B. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We have two identical wires A and B made of different materials. The density of wire A (ρ_A) is greater than the density of wire B (ρ_B), and both wires have the same length (L).

### Step 2: Identify the Wave Pulse
Identical wave pulses are produced at the bottom of both wires. We need to find the time taken for these pulses to reach the top of each wire.

### Step 3: Determine Tension in the Wires
The tension (T) in a wire at a distance x from the bottom can be expressed as:
\[ T = m \cdot g \]
Where \( m \) is the mass of the wire segment above the point at distance x, and \( g \) is the acceleration due to gravity.

The mass can be calculated as:
\[ m = \text{density} \times \text{cross-sectional area} \times \text{length} \]
Thus, for wire A:
\[ T_A = \rho_A \cdot A \cdot x \cdot g \]
And for wire B:
\[ T_B = \rho_B \cdot A \cdot x \cdot g \]

### Step 4: Calculate the Linear Mass Density (μ)
The linear mass density (μ) for each wire is given by:
\[ \mu = \frac{m}{L} = \frac{\rho \cdot A}{L} \]
Thus for wire A:
\[ \mu_A = \frac{\rho_A \cdot A}{L} \]
And for wire B:
\[ \mu_B = \frac{\rho_B \cdot A}{L} \]

### Step 5: Determine the Wave Velocity
The velocity (v) of a wave pulse in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
For wire A:
\[ v_A = \sqrt{\frac{T_A}{\mu_A}} = \sqrt{\frac{\rho_A \cdot A \cdot x \cdot g}{\frac{\rho_A \cdot A}{L}}} = \sqrt{x \cdot g \cdot L} \]

For wire B:
\[ v_B = \sqrt{\frac{T_B}{\mu_B}} = \sqrt{\frac{\rho_B \cdot A \cdot x \cdot g}{\frac{\rho_B \cdot A}{L}}} = \sqrt{x \cdot g \cdot L} \]

### Step 6: Compare Velocities
From the calculations, we see that:
\[ v_A = v_B = \sqrt{x \cdot g \cdot L} \]
This indicates that the wave pulse travels at the same speed in both wires.

### Step 7: Calculate Time Taken
The time taken (t) for the wave pulse to travel the length L is given by:
\[ t = \frac{L}{v} \]
Since \( v_A = v_B \):
\[ t_A = t_B = \frac{L}{\sqrt{g \cdot L}} = \sqrt{\frac{L}{g}} \]

### Conclusion
The time taken by the wave pulse to reach the top of both wires A and B is the same.

### Final Answer
The time taken by the pulse to reach the top is the same for both wires.

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