Each of the two strings of length 51.6 ~cm and 49.1 ~cm are...

Key Idea The number of beats will be the difference of frequencies of the two strings. Frequency of first string $f_1=\frac{1}{2 l_1} \sqrt{\frac{T}{m}}$$\begin{aligned}& =\frac{1}{2 \times 51.6 \times 10^{-2}} \sqrt{\frac{20}{10^{-3}}}\\ & =137.03 \mathrm{~Hz} \end{aligned}$$$Similarly, frequency of second string$$\begin{aligned}& =\frac{1}{2 \times 49.1 \times 10^{-2}} \sqrt{\frac{20}{10^{-3}}}\\ & =144.01 \end{aligned}$$Number$ of beats=\mathrm{f}_2-\mathrm{f}_1=144-137=7 beats