Two strings (A, B) having linear densities $$\mu_{A}=2\times10^{-4}kg/m\text{ and },\mu_{B}=4\times10^{-4}kg/m$$ and lengths $$L_{A}=2.5m$$ and $$L_{B}=1.5m$$ 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 $$t_{1}\text{ and } t_{2}$$, respectively, to reach the joint. The ratio $$t_{1}/ t_{2}$$ is :

String A: $$\mu_A = 2\times10^{-4}$$ kg/m, $$L_A = 2.5$$ m. String B: $$\mu_B = 4\times10^{-4}$$ kg/m, $$L_B = 1.5$$ m. Tension T = 500 N.

Wave speed: $$v = \sqrt{T/\mu}$$.

$$v_A = \sqrt{500/(2\times10^{-4})} = \sqrt{2.5\times10^6} = 500\sqrt{10}$$ m/s

$$v_B = \sqrt{500/(4\times10^{-4})} = \sqrt{1.25\times10^6} = 500\sqrt{5}$$ m/s

$$t_1 = L_A/v_A = 2.5/(500\sqrt{10})$$

$$t_2 = L_B/v_B = 1.5/(500\sqrt{5})$$

$$t_1/t_2 = \frac{2.5}{\sqrt{10}} \times \frac{\sqrt{5}}{1.5} = \frac{2.5\sqrt{5}}{1.5\sqrt{10}} = \frac{2.5}{1.5\sqrt{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx \frac{7.07}{6} \approx 1.18$$

The answer is Option 4: 1.18.