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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2011 · Paper 2 · Physics · Waves and Sound

Column I showns four systems, each of the same length L , for...

Column I showns four systems, each of the same length LL, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as λf\lambda_f. Match each system with statements given in Column II describing the nature and wave length of the standing waves. Column I showns four systems, each of the same length L , for producing standing waves. The lowest possible natural

Options

  1. A(A) p,t, (B) p,s, (C) q,s, (D) q,r
  2. B(A) p,q,t, (B) p,r, (C) q,r, (D) q,s
  3. C(A) p,s, (B) p,r, (C) q,s, (D) q,r
  4. D(A) p,q,r,t, (B) p,s, (C) q,r, (D) q,s

Correct answer

A. (A) p,t, (B) p,s, (C) q,s, (D) q,r

Step-by-step solution

In organ pipes, longitudinal waves are formed. In string, transverse waves are formed. Open end of pipe is displacement antinode and closed end is displacement node. In case of string fixed end of a string is node. Further, least distance between a node and an antinode is λ4\frac{\lambda}{4} and between two nodes is λ2\frac{\lambda}{2}. Keeping these points in mind answer to this question is as under; (A)→(p,t)(\mathrm{A}) \rightarrow(\mathrm{p}, \mathrm{t}) (B) →(p,s)\rightarrow(\mathrm{p}, \mathrm{s}) (C)→(q,s)(\mathrm{C}) \rightarrow(\mathrm{q}, \mathrm{s}) (D) →\rightarrow (q, r) Analysis of Question (i) Unlike the previous years's few match the column problems, this question seems to be more logical. If you know the subject, you can select correct matching without confusion. (ii) In option (d) of column I, string is fixed from centre, therefore at lowest frequency two loops will be formed as shown below. Column I showns four systems, each of the same length L , for producing standing waves. The lowest possible natural
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