JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Waves & Sound

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✓ Correct

Answer: (D)

Solution

The pulse speed is

v = T / μ

, so the travel time along a string is

t = \frac{L}{T / μ} = L μ / T

. The required ratio is

\frac{t _{1}}{t _{2}} = \frac{L _{A}}{L _{B}} \frac{μ _{A}}{μ _{B}} = \frac{2.5}{1.5} \frac{2 \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 4}} = \frac{5}{3} \cdot \frac{1}{2} = \frac{5 2}{6} \approx 1.18

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Waves & Sound

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