1. Formula for fundamental frequency of a stretched wire

For a sonometer wire of length $L$, the fundamental frequency is

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where:

• $T$ = tension in wire
• $\mu$ = mass per unit length

Since $\mu = \rho A$ and $T = YA \times \text{strain}$,

$\frac{T}{\mu} = \frac{YA(\text{strain})}{\rho A} = \frac{Y(\text{strain})}{\rho}$

So,

$f = \frac{1}{2L}\sqrt{\frac{Y\,e}{\rho}}$

where $e$ is the strain.

2. Given data

• $L = 1.5\,m$
• Strain $e = 1\% = 0.01$
• $Y = 2.2 \times 10^{11}\,N/m^2$
• $\rho = 7.7 \times 10^3\,kg/m^3$

Substitute into the formula:

$f = \frac{1}{2(1.5)}\sqrt{\frac{2.2 \times 10^{11} \times 0.01}{7.7 \times 10^3}}$

3. Simplify inside the square root

$2.2 \times 10^{11} \times 0.01 = 2.2 \times 10^9$

Thus,

$\frac{2.2 \times 10^9}{7.7 \times 10^3} = \frac{2.2}{7.7} \times 10^6$

$= 0.2857 \times 10^6 = 2.857 \times 10^5$

Now,

$\sqrt{2.857 \times 10^5} \approx 534.5$

4. Calculate frequency

$f = \frac{1}{3} \times 534.5 \approx 178.2\,Hz$

5. Option check

• A: $188.5\,Hz$ ❌
• B: $178.2\,Hz$ ✅
• C: $200.5\,Hz$ ❌
• D: $770\,Hz$ ❌

Therefore, the correct answer is:

$\boxed{178.2\,Hz}$

So, Option B is correct.