Q115 / 180 · NEET 2013 (All India) · Physics

A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency $250 \mathrm{~Hz}$. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 \mathrm{~Hz}$. The unknown frequency is

  • Correct answer
    A
    $254 \mathrm{~Hz}$
  • B
    $246 \mathrm{~Hz}$
  • C
    $240 \mathrm{~Hz}$
  • D
    $260 \mathrm{~Hz}$

Solution

Here, Figure
We know beat frequency is the difference of the frequencies of the sources. When unknown source is sounded with known source of frequency 250 Hz, it gives 4 beats/s. It means the frequency of unknown source may be 254 Hz or 246 Hz. Now second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. It means the frequency of unknown source may be 518 Hz or 508 Hz. \(\text {As } \frac{518}{2}=259 \mathrm{~Hz} \text { and } \frac{508}{2}=254 \mathrm{~Hz}\) Hence unknown frequency is 254 Hz.

Hence, the correct answer is option: 'A'.

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