When an open pipe is vertically dipped in water and when 8 cm is inside the water then the fundamental frequency of air column is 400 Hz. If it is 6 cm inside the water, then fundamental frequency of the air column is 300 Hz. The length of the tube is : (neglect end correction)

22 cm

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20 cm

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14 cm

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16 cm

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Solution

The correct option is B 14 cm
Frequency, $f \propto \frac{1}{l e n g t h}$
$\frac{f_{1}}{f_{2}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{400}{300} = \frac{l - 6}{l - 8}$
$4 l - 32 = 3 l - 18$
$l = 14 c m$

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When an open pipe is vertically dipped in water and when 8 cm is inside the water then the fundamental frequency of air column is 400 Hz. If it is 6 cm inside the water, then fundamental frequency of the air column is 300 Hz. The length of the tube is : (neglect end correction)

The correct option is B 14 cmFrequency, f∝1lengthf1f2=l2l1⇒400300=l−6l−84l−32=3l−18l=14cm

The correct option is B 14 cm
Frequency, $f \propto \frac{1}{l e n g t h}$
$\frac{f_{1}}{f_{2}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{400}{300} = \frac{l - 6}{l - 8}$
$4 l - 32 = 3 l - 18$
$l = 14 c m$