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NEETPhysicsWaves and Sound

An open organ pipe has a fundamental frequency f . It is dipped vertically into water such that a certain fraction of its length is submerged. Match the submerged fraction in List-I with the ratio of the new fundamental frequency to the original fundamental frequency ( f'/f ) in List-II. List-I (Submerged fraction) List-II ( f'/f ) (A) 1/2 (I) 1 (B) 3/4 (II) 2 (C) 1/3 (III) 3/4 (D) 1/4 (IV) 2/3 Choose the correct ans

Options

  1. A(A) (I), (B) (II), (C) (III), (D) (IV)
  2. B(A) (II), (B) (I), (C) (IV), (D) (III)
  3. C(A) (I), (B) (III), (C) (II), (D) (IV)
  4. D(A) (II), (B) (IV), (C) (I), (D) (III)

Correct answer

A. (A) (I), (B) (II), (C) (III), (D) (IV)

Step-by-step solution

Let the original length of the open pipe be L . Its fundamental frequency is f = v 2L . When a fraction x of the pipe is submerged, the length of the air column remaining is L' = L(1 - x) . The pipe now acts as a closed organ pipe. Its new fundamental frequency is f' = v 4L' = v 4L(1 - x) . The ratio of the new frequency to the original frequency is: f' f = v 4L(1 - x) v 2L = 1 2(1 - x) For (A) x = 1/2 : f' f = 1 2(1 - 1/2) = 1 For (B) x = 3/4 : f' f = 1 2(1 - 3/4) = 2 For (C) x = 1/3 : f' f = 1 2(1 - 1/3) = 3 4 Fo

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