MCQEasyJEE Main 2026 · 21 January, Shift 1Wave Motion Basics

Physics Question from JEE Main 2026 · 21 January, Shift 1

Two strings (AA, BB) having linear densities μA=2×104kg/m\mu_A = 2 \times 10^{-4} \, \text{kg/m} and, μB=4×104kg/m\mu_B = 4 \times 10^{-4} \, \text{kg/m} and lengths LA=2.5mL_A = 2.5 \, \text{m} and LB=1.5mL_B = 1.5 \, \text{m} respectively are joined. Free ends of AA and BB are tied to two rigid supports CC and DD, respectively creating a tension of 500N500 \, \text{N} in the wire. Two identical pulses, sent from CC and DD ends, take time tAt_A and tBt_B, respectively, to reach the joint. The ratio tA/tBt_A/t_B is:

  • A

    1.081.08

  • B

    1.901.90

  • C

    1.181.18

  • D

    1.671.67

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