1. Relation for frequency of a sonometer wire

For a given wire under the same tension and same linear density, the fundamental frequency is

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

So, for fixed $T$ and $\mu$,

$f \propto \frac{1}{L}$

2. Given condition of beats

A tuning fork produces $5$ beats per second with the sonometer wire when its length is either $0.95\,\text{m}$ or $1.0\,\text{m}$.

That means the tuning fork frequency lies between the two corresponding sonometer frequencies, and differs by $5\,\text{Hz}$ from each:

Let the sonometer frequencies at lengths $0.95\,\text{m}$ and $1.0\,\text{m}$ be $f_1$ and $f_2$ respectively.

Since $f \propto 1/L$,

$\frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{1.0}{0.95} = \frac{20}{19}$

Also, shorter length gives higher frequency, so

$f_1 > f_2$

If the fork gives 5 beats with both, then the fork frequency $f$ is between them:

$f_1 - f = 5, \qquad f - f_2 = 5$

Thus,

$f_1 - f_2 = 10$

3. Use the ratio and difference

Let

$f_2 = x \implies f_1 = \frac{20}{19}x$

Now,

$f_1 - f_2 = 10$

$\frac{20}{19}x - x = 10$

$\left(\frac{20}{19} - 1\right)x = 10$

$\frac{1}{19}x = 10$

$x = 190$

So,

$f_2 = 190\,\text{Hz}, \qquad f_1 = 200\,\text{Hz}$

4. Find the tuning fork frequency

Since the tuning fork is midway between them:

$f = f_2 + 5 = 190 + 5 = 195\,\text{Hz}$

5. Check options

• A: $195\,\text{Hz}$ ✅
• B: $150\,\text{Hz}$ ❌
• C: $300\,\text{Hz}$ ❌
• D: $251\,\text{Hz}$ ❌

Therefore, the correct answer is:

$\boxed{195\,\text{Hz}}$