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JEE Main 2019 (9 Apr, Shift 1) · Speed of Waves

Speed of Waves question from JEE Main 2019 (9 Apr, Shift 1)

Q33JEE Main 2019 · 9 APR · SHIFT 1Difficult

Speed of Waves

The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1. Approximate value of T is

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Answer: (c)

  1. At 0∘C0^\circ\text{C}: v0=ωk=10003=333.3 m/sv_0 = \dfrac{\omega}{k} = \dfrac{1000}{3} = 333.3\text{ m/s}.
  2. v∝Tv \propto \sqrt{T}, so T=273(336333.3)2≈273×1.016≈277.4 KT = 273\left(\dfrac{336}{333.3}\right)^2 \approx 273 \times 1.016 \approx 277.4\text{ K}
  3. T≈4.4∘CT \approx 4.4^\circ\text{C}
  4. So the answer is (c) 4∘C4^\circ\text{C}.
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