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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

2

b

1

c

23

d

32

answer is B.

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Detailed Solution

For open tube, n0=v2l
For 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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