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NEET PHYSICSWAVE OPTICSMedium

Question

If n1,n2n_1, n_2 and n3n_3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency nn of the string is given by

A

1n=1n1+1n2+1n3\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}

B

1n=1n1+1n2+1n3\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_1}}+\frac{1}{\sqrt{n_2}}+\frac{1}{\sqrt{n_3}}

C

n=n1+n2+n3\sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}

D

n=n1+n2+n3n=n_1+n_2+n_3

Step-by-Step Solution

  1. Identify the formula for fundamental frequency: The fundamental frequency nn of a stretched string of length ll, tension TT, and linear mass density μ\mu is given by n=12lTμn = \frac{1}{2l}\sqrt{\frac{T}{\mu}} .
  2. Relate length to frequency: Assuming the tension TT and linear mass density μ\mu remain constant for all segments of the string, the length ll is inversely proportional to the frequency nn. Thus, l=Cnl = \frac{C}{n}, where C=12TμC = \frac{1}{2}\sqrt{\frac{T}{\mu}} is a constant.
  3. Use the property of total length: The total length of the string is equal to the sum of the lengths of the individual segments into which it is divided: l=l1+l2+l3l = l_1 + l_2 + l_3
  4. Substitute the length-frequency relationship: Substituting l=Cnl = \frac{C}{n} into the length equation gives: Cn=Cn1+Cn2+Cn3\frac{C}{n} = \frac{C}{n_1} + \frac{C}{n_2} + \frac{C}{n_3} Dividing both sides by the constant CC, we obtain the relationship between the fundamental frequencies: 1n=1n1+1n2+1n3\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}

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 OPTICSfundamentalfrequenciessegmentsstringdivided

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