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NEET PHYSICSWAVE OPTICSMedium

Question

A source of unknown frequency gives 4 beats/s4 \text{ beats/s} when sounded with a source of known frequency 250 Hz250 \text{ Hz}. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 513 Hz513 \text{ Hz}. The unknown frequency is

A

254 Hz254 \text{ Hz}

B

246 Hz246 \text{ Hz}

C

240 Hz240 \text{ Hz}

D

260 Hz260 \text{ Hz}

Step-by-Step Solution

  1. Determine possible frequencies from the first condition: The beat frequency is the difference between the two frequencies. Let the unknown frequency be ff. Since it produces 4 beats/s with a source of 250 Hz250 \text{ Hz} , f250=4|f - 250| = 4 f=250±4f = 250 \pm 4 f=254 Hz or 246 Hzf = 254 \text{ Hz} \text{ or } 246 \text{ Hz}
  2. Determine possible frequencies from the second condition: The second harmonic of the unknown frequency is 2f2f. This produces 5 beats/s with a source of 513 Hz513 \text{ Hz}. 2f513=5|2f - 513| = 5 2f=513±52f = 513 \pm 5 2f=518 Hz or 508 Hz2f = 518 \text{ Hz} \text{ or } 508 \text{ Hz} f=259 Hz or 254 Hzf = 259 \text{ Hz} \text{ or } 254 \text{ Hz}
  3. Find the common frequency: Comparing the possible values of ff from both conditions, the only common value is f=254 Hzf = 254 \text{ Hz}. Therefore, the unknown frequency is 254 Hz254 \text{ Hz}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSsourceunknownfrequencybeatsssounded

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