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Waves question

2012 · Shift 0 · Q61
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Waves question

2012 · Shift 0 · Q61

JEE MainPhysicsWavesMCQ+4 / −1
A cylindrical tube, open at both ends, has a fundamental frequency, f,f,f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now :
  1. A
    fff
  2. B
    f/2f/2f/2
  3. C
    3/43/43/4
  4. D
    2f2f2f
View written solutionFree

Correct answer: A

  1. Initial tube: open at both ends

For a tube of length LLL open at both ends, the fundamental frequency is

f=v2Lf = \frac{v}{2L}f=2Lv​

where vvv is the speed of sound in air.

  1. After dipping half the tube in water

If half the tube is dipped vertically in water, then the lower half is filled with water and only the upper half contains air.

So the new air-column length is

L′=L2L' = \frac{L}{2}L′=2L​

Also, the water surface acts like a closed end, while the top remains open. Thus the tube now behaves like a closed pipe of length L/2L/2L/2.

  1. Fundamental frequency of closed pipe

For a pipe closed at one end and open at the other, the fundamental frequency is

f′=v4L′f' = \frac{v}{4L'}f′=4L′v​

Substitute L′=L/2L' = L/2L′=L/2:

f′=v4(L/2)=v2Lf' = \frac{v}{4(L/2)} = \frac{v}{2L}f′=4(L/2)v​=2Lv​

But originally,

f=v2Lf = \frac{v}{2L}f=2Lv​

Hence,

f′=ff' = ff′=f

  1. Option check
  • A: fff ✅
  • B: f/2f/2f/2 ❌
  • C: 3/43/43/4 ❌ (dimensionally incomplete and incorrect)
  • D: 2f2f2f ❌

Therefore, the fundamental frequency remains unchanged.

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