A wire of mass 5 gram is kept stretched by a force of 400 N. When plucked at a point, transverse waves travel along the wire with a speed of 400 m/s. The length of the wire is

Options

A. 1 m
B. 2 m
C. 3 m
D. 1.5 m

Answer

To find the length of the wire, we can use the formula for the speed of transverse waves in a stretched string. The speed (V) of the wave is given by the equation:

\[ V = \sqrt{\frac{T}{\mu}} \]

where:
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire, defined as \( \mu = \frac{m}{L} \),
- \( m \) is the mass of the wire,
- \( L \) is the length of the wire.

### Step 1: Identify the given values
- Mass of the wire, \( m = 5 \, \text{grams} = 5 \times 10^{-3} \, \text{kg} \)
- Tension in the wire, \( T = 400 \, \text{N} \)
- Speed of the wave, \( V = 400 \, \text{m/s} \)

### Step 2: Substitute the expression for \( \mu \) into the wave speed formula
From the definition of linear mass density:
\[ \mu = \frac{m}{L} \]

Substituting this into the wave speed formula gives:
\[ V = \sqrt{\frac{T}{\frac{m}{L}}} \]

This can be rewritten as:
\[ V = \sqrt{\frac{T \cdot L}{m}} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides results in:
\[ V^2 = \frac{T \cdot L}{m} \]

### Step 4: Rearrange the equation to solve for \( L \)
Rearranging the equation to isolate \( L \):
\[ L = \frac{V^2 \cdot m}{T} \]

### Step 5: Substitute the known values into the equation
Now substituting the known values:
- \( V = 400 \, \text{m/s} \)
- \( m = 5 \times 10^{-3} \, \text{kg} \)
- \( T = 400 \, \text{N} \)

We get:
\[ L = \frac{(400)^2 \cdot (5 \times 10^{-3})}{400} \]

### Step 6: Simplify the expression
Calculating \( (400)^2 = 160000 \):
\[ L = \frac{160000 \cdot (5 \times 10^{-3})}{400} \]

Now simplifying:
\[ L = \frac{800}{400} = 2 \, \text{meters} \]

### Final Answer
The length of the wire is \( L = 2 \, \text{meters} \).

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