A cylindrical tube of length 150 cm open at both ends has fundamental frequency $f_{0}$ in air. Now the tube is dipped vertically in a liquid so that 100 cm is still out of the liquid. Find fundamental frequency of air column (new).

2 f_{0}

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\frac{f_{0}}{2}

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\frac{f_{0}}{3}

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\frac{3 f_{0}}{4}

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Solution

The correct option is D $\frac{3 f_{0}}{4}$
Initially, air column in a pipe open at both ends
$f_{0} = \frac{v}{2 l} \dots \dots (1)$
$\frac{v}{2 (1.5)} = \frac{v}{3}$

For air column in a pipe closed at one end,
$f_{0}^{^{'}} = \frac{v}{4 (1)} = \frac{v}{4} = \frac{3 f_{0}}{4}$

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A cylindrical tube of length 150 cm open at both ends has fundamental frequency f0 in air. Now the tube is dipped vertically in a liquid so that 100 cm is still out of the liquid. Find fundamental frequency of air column (new).

The correct option is D 3f04 Initially, air column in a pipe open at both ends f0=v2l……(1) v2(1.5)=v3 For air column in a pipe closed at one end, f′0=v4(1)=v4=3f04

A cylindrical tube of length 150 cm open at both ends has fundamental frequency $f_{0}$ in air. Now the tube is dipped vertically in a liquid so that 100 cm is still out of the liquid. Find fundamental frequency of air column (new).

The correct option is D $\frac{3 f_{0}}{4}$
Initially, air column in a pipe open at both ends
$f_{0} = \frac{v}{2 l} \dots \dots (1)$
$\frac{v}{2 (1.5)} = \frac{v}{3}$

For air column in a pipe closed at one end,
$f_{0}^{^{'}} = \frac{v}{4 (1)} = \frac{v}{4} = \frac{3 f_{0}}{4}$