A cylindrical tube, open at both ends, has a fundamental frequency $f_{0}$ in air.

The tube is dipped vertically into water such that half of its length is inside

water. The fundamental frequency of the air column now is

[RPET 1999; RPMT 1998, 2000; J & K CET 2000;

KCET 2002; BHU 2002; BCECE 2003]

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Solution

The correct option is B

$n_{o p e n} = \frac{v}{2 l_{o p e n}}$

$n_{c l o s e d} = \frac{v}{4 l_{c l o e s}} = \frac{v}{\frac{4 l_{o p e n}}{2}} = \frac{v}{2 l_{o p e n}}$

$(A s l_{c l o s e d} = \frac{l_{o p e n}}{2})$ , i.e. frequency remains unchanged.

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A cylindrical tube, open at both ends, has a fundamental frequency f0 in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is [RPET 1999; RPMT 1998, 2000; J & K CET 2000; KCET 2002; BHU 2002; BCECE 2003]

The correct option is B nopen=v2lopen nclosed=v4lcloes=v4lopen2=v2lopen (As lclosed=lopen2) , i.e. frequency remains unchanged.

The correct option is B

$n_{o p e n} = \frac{v}{2 l_{o p e n}}$

$n_{c l o s e d} = \frac{v}{4 l_{c l o e s}} = \frac{v}{\frac{4 l_{o p e n}}{2}} = \frac{v}{2 l_{o p e n}}$

$(A s l_{c l o s e d} = \frac{l_{o p e n}}{2})$ , i.e. frequency remains unchanged.