JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

Two strings (A, B) having linear densities $μ_{A} = 2 \times 1 0^{- 4}$ kg/m and $μ_{B} = 4 \times 1 0^{- 4}$ 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 :

A
1.08
B
1.90
C
1.67
D
1.18

Solution

Tension is the same along the wire (= 500 N). Wave speed: $v = T / μ$ , so the ratio $v_{A} / v_{B} = μ_{B} / μ_{A} = 2$ .

Travel times: $t_{1} = L_{A} / v_{A}$ and $t_{2} = L_{B} / v_{B}$ . Therefore

$\frac{t _{1}}{t _{2}} = \frac{L _{A}}{L _{B}} \cdot \frac{v _{B}}{v _{A}} = \frac{2.5}{1.5} \cdot \frac{1}{2} = \frac{5}{3 2} \approx 1.18$ .

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