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JEE Main 2019 (12 Jan, Shift 2) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from JEE Main 2019 (12 Jan, Shift 2)

Q105JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Organ Pipes & Resonance Tube

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to:

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Answer: (b)

  1. First resonance: l+e=λ4=v4fl + e = \frac{\lambda}{4} = \frac{v}{4f}; the unknown end/reference offset ee is the same in both runs
  2. Subtract: l2−l1=v4(1256−1512)=v2048l_2 - l_1 = \frac{v}{4}\left(\frac{1}{256} - \frac{1}{512}\right) = \frac{v}{2048}
  3. 0.27−0.11=0.16=v2048⇒v=327.7 m/s0.27 - 0.11 = 0.16 = \frac{v}{2048} \Rightarrow v = 327.7\text{ m/s}
  4. So the answer is (b) 328 m s−1328\text{ m s}^{-1}.
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