Given: Lengths of the strings, $l_1 = 51.6 \text{ cm} = 0.516 \text{ m}$ and $l_2 = 49.1 \text{ cm} = 0.491 \text{ m}$ Tension in the strings, $T = 20 \text{ N}$ Mass per unit length, $\mu = 1 \text{ g/m} = 1 \times 10^{-3} \text{ kg/m}$

The velocity of a transverse wave in a stretched string is given by: $v = \sqrt{\frac{T}{\mu}}$ $v = \sqrt{\frac{20}{1 \times 10^{-3}}} = \sqrt{20000} \approx 141.42 \text{ m/s}$

The fundamental frequency of a vibrating string of length $l$ is $f = \frac{v}{2l}$. The frequencies of the two strings are: $f_1 = \frac{v}{2l_1} = \frac{141.42}{2 \times 0.516} = \frac{141.42}{1.032} \approx 137.03 \text{ Hz}$ $f_2 = \frac{v}{2l_2} = \frac{141.42}{2 \times 0.491} = \frac{141.42}{0.982} \approx 144.01 \text{ Hz}$

The number of beats produced per second is the difference between the two frequencies: $\text{Number of beats} = f_2 - f_1 = 144.01 - 137.03 = 6.98 \approx 7$