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 answerA$254 \mathrm{~Hz}$
- B$246 \mathrm{~Hz}$
- C$240 \mathrm{~Hz}$
- D$260 \mathrm{~Hz}$
Solution
Here,

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.

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'.