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JEE Main 2020 (9 Jan, Shift 2) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from JEE Main 2020 (9 Jan, Shift 2)

Q83JEE Main 2020 · 9 JAN · SHIFT 2Medium

Stationary Waves & Vibrating Strings

A wire of length L and mass per unit length 6.0 × 10–3 kgm–1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is:

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Answer: (b)

  1. Fundamental n1=490−420=70 Hzn_1 = 490 - 420 = 70\text{ Hz}.
  2. v=T/μ=5406×10−3=300 m/sv = \sqrt{T/\mu} = \sqrt{\dfrac{540}{6\times10^{-3}}} = 300\text{ m/s}
  3. L=v2n1=300140≈2.14 mL = \dfrac{v}{2n_1} = \dfrac{300}{140} \approx 2.14\text{ m}
  4. So the answer is (b) 2.1 m.
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