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← IIT-JEE 2010 · Paper 2
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IIT-JEE 2010 · Paper 2, Physics question 40

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 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 N and the speed of sound is 320 m s⁻¹, the mass of the string is

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Answer and worked solution

Official answer: B

Pipe (closed at one end): fundamental f=v4L=3203.2=100f=\dfrac v{4L}=\dfrac{320}{3.2}=100 Hz.

String in its second harmonic: f=2⋅vs2l=vs0.5=100f=2\cdot\dfrac{v_s}{2l}=\dfrac{v_s}{0.5}=100 Hz, so vs=50v_s=50 m s⁻¹.

Mass. vs=T/μv_s=\sqrt{T/\mu} gives μ=502500=0.02\mu=\dfrac{50}{2500}=0.02 kg m⁻¹, and the 0.5 m string has mass 0.010.01 kg =10=10 g.

(B)

Topics:Standing waves in stringsOrgan pipes and resonance
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