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MHT CET 2026 (17 Apr, Shift 2) · Beats

Beats question from MHT CET 2026 (17 Apr, Shift 2)

Q8MHT CET 2026 · 17 APR · SHIFT 2Medium

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Two sounding waves send waves at certain temperature in air of wavelength 50 cm and 50.5 cm repectively. The frequency of sources differ by 6 Hz. The velocity of sound in air at same temperature is

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Answer: (c)

  1. n1−n2=vλ1−vλ2=6n_1 - n_2 = \dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2} = 6 with λ1=0.50 m\lambda_1 = 0.50\text{ m}, λ2=0.505 m\lambda_2 = 0.505\text{ m}.
  2. v⋅0.505−0.500.50×0.505=6⇒v⋅0.0050.2525=6v \cdot \dfrac{0.505 - 0.50}{0.50 \times 0.505} = 6 \Rightarrow v \cdot \dfrac{0.005}{0.2525} = 6
  3. v=6×0.25250.005=303 m/sv = \dfrac{6 \times 0.2525}{0.005} = 303\text{ m/s}
  4. So the answer is (c) 303 m/s.
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