A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency $250…
- $254 \mathrm{~Hz}$
- $246 \mathrm{~Hz}$
- $240 \mathrm{~Hz}$
- $260 \mathrm{~Hz}$
Solution

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.
Asked in: NEET 2013 (All India)