Q87 / 90 · JEE Main 2022 (27 Jul Shift 2) · Physics

A wire of length 30 cm, stretched between rigid supports, has it's nth  and n+1th harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is _____ kg m-1.

Correct answer: 3

Solution

Frequency of nth harmonic is nv2l=400⇒nv0.6=400   ...1 

And frequency of n+1th harmonic is n+1v2l=450⇒n+1v0.6=450    ...2, here, v is speed of wave on string. 

From equation 1, we have

n=400×0.6v

Putting in equation 2, 

⇒0.6×400v+1v0.6=450

⇒400+v0.6=450

⇒v=30 m s-1

Speed of wave on string is stated as v=Tμ=30, here, μ is linear mass density.

So, μ=Tv2=2700900⇒μ=3 kg m-1

Hence, the correct answer is 3.

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