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JEE Main 2025 (8 Apr, Shift 2) · Speed of Waves

Speed of Waves question from JEE Main 2025 (8 Apr, Shift 2)

Q11JEE Main 2025 · 8 APR · SHIFT 2Easy

Speed of Waves

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is v1v_1, and that in the other string having radius of cross section R/2 is v2v_2. Then v2v1\frac{v_2}{v_1} =

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Answer: (c)

  1. μ=ρπR2\mu = \rho \pi R^2, so with equal TT: v=T/μ∝1Rv = \sqrt{T/\mu} \propto \dfrac{1}{R}.
  2. v2v1=RR/2=2\dfrac{v_2}{v_1} = \dfrac{R}{R/2} = 2
  3. So the answer is (c) 22.
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