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MHT CET 2025 (19 Apr, Shift 2) · Beats

Beats question from MHT CET 2025 (19 Apr, Shift 2)

Q23MHT CET 2025 · 19 APR · SHIFT 2Medium

Beats

When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same sonometer wire. The frequencies of the two tuning forks are

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Answer: (b)

  1. f∝1Lf \propto \dfrac{1}{L}: f1×0.69=f2×0.70f_1 \times 0.69 = f_2 \times 0.70, with f1−f2=6f_1 - f_2 = 6.
  2. f2(0.700.69−1)=6⇒f2=6×0.690.01=414 Hzf_2\left(\dfrac{0.70}{0.69} - 1\right) = 6 \Rightarrow f_2 = \dfrac{6 \times 0.69}{0.01} = 414\text{ Hz}
  3. f1=420 Hzf_1 = 420\text{ Hz}
  4. So the answer is (b) 414 Hz, 420 Hz.
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