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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Waves and Sound

JEE Advanced 2010 · Paper 2 · Physics · Waves and Sound

A hollow pipe of length 0.8 ~m is closed at one end. At its open...

A hollow pipe of length 0.8 m0.8 \mathrm{~m} is closed at one end. At its open end a 0.5 m0.5 \mathrm{~m} long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N50 \mathrm{~N} and the speed of sound is 320 ms−1320 \mathrm{~ms}^{-1}, the mass of the string is

Options

  1. A5 g5 \mathrm{~g}
  2. B10 g10 \mathrm{~g}
  3. C20 g20 \mathrm{~g}
  4. D40 g40 \mathrm{~g}

Correct answer

B. 10 g10 \mathrm{~g}

Step-by-step solution

2(v12l1)=v24l2∴T/μl1=3204l22\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2} ( μ=\mu= mass per unit length of wire) or 50/μ0.5=3204×0.8\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8} Solving we get μ=0.02kgm=20gm∴\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}\therefore Mass of string =\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5  m)=\mathrm{~m})=10  g∴\mathrm{~g}\therefore The correct option is (b).
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