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JEE Main 201323 AprilOnline
Exam
JEE Main
Year
2013
Date
23 April
Mode
Online
Subject
Physics
Chapter
Waves and Sound

JEE Main 2013 · 23 April · Online · Physics · Waves and Sound

A sonometer wire of length 114 ~cm is fixed at both the ends. Where...

A sonometer wire of length 114 cm114 \mathrm{~cm} is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio 1:3:41: 3: 4 ?

Options

  1. AAt 36 cm36 \mathrm{~cm} and 84 cm84 \mathrm{~cm} from one end
  2. BAt 24 cm24 \mathrm{~cm} and 72 cm72 \mathrm{~cm} from one end
  3. CAt 48 cm48 \mathrm{~cm} and 96 cm96 \mathrm{~cm} from one end
  4. DAt 72 cm72 \mathrm{~cm} and 96 cm96 \mathrm{~cm} from one end

Correct answer

D. At 72 cm72 \mathrm{~cm} and 96 cm96 \mathrm{~cm} from one end

Step-by-step solution

Total length of the wire, L=114 cmn1:n2:n3=1:3:4\mathrm{L}=114 \mathrm{~cm}\mathrm{n}_1: \mathrm{n}_2: \mathrm{n}_3=1: 3: 4 Let L1, L2\mathrm{L}_1, \mathrm{~L}_2 and L3\mathrm{L}_3 be the lengths of the three parts As n∝1 L\mathrm{n} \propto \frac{1}{\mathrm{~L}}∴L1:L2:L3=11:13:14=12:4:3∴L1=72 cm(1212+4+3×114)L2=24 cm(419×114) and  L3=18 cm(319×114)\begin{aligned}& \therefore \quad \mathrm{L}_{1}: \mathrm{L}_{2}: \mathrm{L}_{3}=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3 \\ & \therefore \quad \mathrm{L}_{1}=72 \mathrm{~cm}\left(\frac{12}{12+4+3} \times114\right) \\ & \mathrm{L}_{2}=24 \mathrm{~cm}\left(\frac{4}{19} \times114\right) \\ & \text{\text{ and } } \quad \mathrm{L}_{3}=18 \mathrm{~cm}\left(\frac{3}{19} \times114\right) \end{aligned}HenceHence the bridges should be placed at72 \mathrm{~cm} and 72+24=96 \mathrm{~cm} from one end.
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