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Q.

An open tube is in resonance with string (frequency of vibration of tube is n0). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be

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a

1

b

2

c

23

d

32

answer is B.

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Detailed Solution

For open tube, n0=v2lFor closed tube length available for resonance is l'=l×25100=l4 ∴Fundamental frequency of water filled tube n=v4l'=v4×(l/4)=vl=2n0⇒nn0=2

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