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NEET PhysicsGeneralHard

Question

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:

1

523 Hz

2

524 Hz

3

536 Hz

4

537 Hz

Step-by-Step Solution

Difference of fAf_A and fBf_B is 6 Hz. If tension decreases, fBf_B decreases to fBf_B'. New difference fAfB=7 Hz|f_A - f_B'| = 7 \text{ Hz}. Since the beat frequency increased, fA>fBf_A > f_B. Thus fAfB=6 Hz    530fB=6    fB=524 Hzf_A - f_B = 6 \text{ Hz} \implies 530 - f_B = 6 \implies f_B = 524 \text{ Hz}.

Exam Context & Concepts Covered

This question aligns with the NEET Physics syllabus, specifically targeting concepts from General. 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.

Physicsguitarstringsmaterialslightlyproduce

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