When tension of a string is increased by 2.5 N, the initial frequency is altered in the ratio of 3:2. The initial tension in the string is

Options

A. 6N
B. 5N
C. 4N
D. 2N

Explanation

To solve the problem, we need to find the initial tension in the string given that when the tension is increased by 2.5 N, the frequency changes in the ratio of 3:2.

### Step-by-Step Solution:

1. **Define Variables**:
Let the initial tension in the string be \( T \) (in Newtons). The initial frequency is denoted as \( f \).

2. **Frequency Relation**:
The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the string and \( \mu \) is the linear mass density of the string.

3. **New Tension and Frequency**:
When the tension is increased by 2.5 N, the new tension becomes \( T + 2.5 \) N. The new frequency is given as:
\[ f' = \frac{1}{2L} \sqrt{\frac{T + 2.5}{\mu}} \]

4. **Frequency Ratio**:
According to the problem, the ratio of the new frequency to the initial frequency is:
\[ \frac{f'}{f} = \frac{3}{2} \]
Substituting the expressions for \( f' \) and \( f \):
\[ \frac{\frac{1}{2L} \sqrt{\frac{T + 2.5}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T}{\mu}}} = \frac{3}{2} \]
This simplifies to:
\[ \frac{\sqrt{T + 2.5}}{\sqrt{T}} = \frac{3}{2} \]

5. **Squaring Both Sides**:
Squaring both sides to eliminate the square roots:
\[ \frac{T + 2.5}{T} = \left(\frac{3}{2}\right)^2 \]
This gives:
\[ \frac{T + 2.5}{T} = \frac{9}{4} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 4(T + 2.5) = 9T \]

7. **Expanding and Rearranging**:
Expanding the left side:
\[ 4T + 10 = 9T \]
Rearranging gives:
\[ 10 = 9T - 4T \]
Thus:
\[ 10 = 5T \]

8. **Solving for T**:
Dividing both sides by 5:
\[ T = 2 \text{ N} \]

### Final Answer:
The initial tension in the string is \( T = 2 \text{ N} \).

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