Q59 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
Two strings $(A, B)$ having linear densities $\mu_{A}=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and, $\mu_{B}=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and lengths $L_{A}=2.5 \mathrm{~m}$ and $L_{B}=1.5 \mathrm{~m}$ 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 :
- A1.90
- B1.67
- C1.08
- Correct answerD1.18
Solution
Wave velocity in string: $v = \sqrt{\frac{T}{\mu}}$
Time to reach joint: $t = \frac{L}{v} = L\sqrt{\frac{\mu}{T}}$
$t_1 = L_A\sqrt{\frac{\mu_A}{T}}$ and $t_2 = L_B\sqrt{\frac{\mu_B}{T}}$
$\frac{t_1}{t_2} = \frac{L_A}{L_B} \times \sqrt{\frac{\mu_A}{\mu_B}} = \frac{2.5}{1.5} \times \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}}$
$= \frac{5}{3} \times \sqrt{\frac{1}{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx 1.18$
Time to reach joint: $t = \frac{L}{v} = L\sqrt{\frac{\mu}{T}}$
$t_1 = L_A\sqrt{\frac{\mu_A}{T}}$ and $t_2 = L_B\sqrt{\frac{\mu_B}{T}}$
$\frac{t_1}{t_2} = \frac{L_A}{L_B} \times \sqrt{\frac{\mu_A}{\mu_B}} = \frac{2.5}{1.5} \times \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}}$
$= \frac{5}{3} \times \sqrt{\frac{1}{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx 1.18$
Hence, the correct answer is option: 'D'.