5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be:

For a sonometer wire vibrating in its fundamental mode the frequency is given by the well-known relation

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

where $$l$$ is the vibrating length, $$T$$ is the tension and $$\mu$$ is the mass per unit length. During the experiment the wire, the tension and the material of the wire remain the same, so the factor $$\sqrt{T/\mu}/2$$ is constant. Let us denote this constant by $$k$$. Hence

$$f=\frac{k}{l}$$

We now write the frequencies of the wire for the two given lengths.

For length $$l_1=0.95\ \text{m}$$, the frequency is

$$f_1=\frac{k}{0.95}$$

For length $$l_2=1\ \text{m}$$, the frequency is

$$f_2=\frac{k}{1}=k$$

The tuning-fork of unknown frequency $$f_f$$ produces 5 beats per second with each of these wire frequencies, so

$$|f_f-f_1|=5 \qquad\text{and}\qquad |f_f-f_2|=5$$

Because $$f_1>f_2$$ (the shorter wire gives the higher note), the only possible way for both moduli to be 5 is for the fork frequency to lie between the two string frequencies, that is

$$f_1-f_f=5 \quad\text{and}\quad f_f-f_2=5$$

Adding these two equalities gives

$$f_1-f_2=10$$

Now compute the difference $$f_1-f_2$$ in terms of $$k$$:

$$f_1-f_2=\frac{k}{0.95}-k=k\!\left(\frac{1}{0.95}-1\right)=k\!\left(\frac{1-0.95}{0.95}\right)=k\!\left(\frac{0.05}{0.95}\right)=k\!\left(\frac{1}{19}\right)$$

We have just found that this difference must equal 10, so

$$k\!\left(\frac{1}{19}\right)=10 \quad\Longrightarrow\quad k=10\times19=190$$

Therefore

$$f_2=k=190\ \text{Hz} \qquad\text{and}\qquad f_1=\frac{190}{0.95}=200\ \text{Hz}$$

The fork frequency, lying midway between $$f_1$$ and $$f_2$$, is

$$f_f=\frac{f_1+f_2}{2}=\frac{200+190}{2}=195\ \text{Hz}$$

Hence, the correct answer is Option A.