Two strings (A, B) having linear densities \(\mu_{ A }=2 \times 10^{-4} kg / m\) and \(\mu_{ B }=4 \times 10^{-4} kg / m\) and lengths \(L _{ A }=2.5 m\) and \(L _{ B }=1.5 m\) respectively are joined. Free ends of A and B are tied to two rigid supports C and D , respectively creating a tension of 500 N in the wire. Two identical pulses, sent from \(C\) and \(D\) ends, take time \(t_1\) and \(t_2\), respectively, to reach the joint. The ratio \(t_1 / t_2\) is :

Given \(L _{ A }=2.5 m\),
\(L _{ B }=1.5 m\),
\(T =500 N\)
\(v_A=\sqrt{\frac{T}{\mu_A}}=\sqrt{\frac{500}{2 \times 10^{-4}}}=5 \sqrt{10} \times 10^2 m / s\)
\(v _{ B }=\sqrt{\frac{ T }{\mu_{ B }}}=\sqrt{\frac{500}{4 \times 10^{-4}}}=5 \sqrt{5} \times 10^2 m / s\)
\(t _1=\frac{ L _{ A }}{ v _{ A }}=\frac{2.5}{5 \sqrt{10}} \times 10^{-2} s\)
\(t_2=\frac{L_B}{v_B}=\frac{1.5}{5 \sqrt{5}} \times 10^{-2} s\)
\(\therefore \frac{ t _1}{ t _2}=\frac{2.5}{5 \sqrt{10}} \times \frac{5 \sqrt{5}}{1.5}=\frac{5}{3} \times \frac{1}{\sqrt{2}}=\frac{1.66}{1.41}=1.18\)