Velocity of a wave in a wire is v when tension in it is `2.06 × 10^4 N`. Find value of tension in wire when velocity of wave become `v/2`.

Options

A. `5.15 xx 10^(3)N`
B. `30.5 xx 10^(4)N`
C. `2.50 xx 10^(4)N`
D. `10.2 xx 10^(2)N`

Answer

To find the tension in the wire when the velocity of the wave becomes \( \frac{v}{2} \), we can follow these steps:

### Step 1: Understand the relationship between wave velocity and tension
The velocity \( v \) of a wave in a wire is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density of the wire. From this, we can see that the velocity is directly proportional to the square root of the tension.

### Step 2: Set up the proportional relationship
When the velocity changes to \( \frac{v}{2} \), we can express this relationship as:
\[ \frac{v'}{v} = \sqrt{\frac{T'}{T}} \]
where \( v' = \frac{v}{2} \) and \( T' \) is the new tension we want to find.

### Step 3: Substitute the new velocity
Substituting \( v' = \frac{v}{2} \) into the equation gives:
\[ \frac{\frac{v}{2}}{v} = \sqrt{\frac{T'}{T}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{T'}{T}} \]

### Step 4: Square both sides
Squaring both sides of the equation results in:
\[ \left(\frac{1}{2}\right)^2 = \frac{T'}{T} \]
which simplifies to:
\[ \frac{1}{4} = \frac{T'}{T} \]

### Step 5: Solve for the new tension
Rearranging the equation gives:
\[ T' = \frac{T}{4} \]

### Step 6: Substitute the known tension
We know the initial tension \( T = 2.06 \times 10^4 \, \text{N} \). Substituting this value into the equation gives:
\[ T' = \frac{2.06 \times 10^4}{4} \]

### Step 7: Calculate the new tension
Calculating this gives:
\[ T' = 5.15 \times 10^3 \, \text{N} \]

### Final Answer
Thus, the value of the tension in the wire when the velocity of the wave becomes \( \frac{v}{2} \) is:
\[ \boxed{5.15 \times 10^3 \, \text{N}} \]

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