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(B) The fundamental frequency of a closed pipe of length $L$ is given by $f_1 = \frac{v}{4L} = 400 \,Hz$.From this,we get $v = 1600L$.When $1/3$ of th

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$1800$

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(B) The fundamental frequency of a closed pipe of length $L$ is given by $f_1 = \frac{v}{4L} = 400 \,Hz$.From this,we get $v = 1600L$.When $1/3$ of the pipe is filled with water,the length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.The new fundamental frequency of the air column is $f'_1 = \frac{v}{4L'} = \frac{v}{4(2L/3)} = \frac{3v}{8L}$.Substituting $v = 1600L$,we get $f'_1 = \frac{3(1600L)}{8L} = 3 	imes 200 = 600 \,Hz$.In a closed pipe,the harmonics are odd multiples of the fundamental frequency $(f_1, 3f_1, 5f_1, \dots)$. However,the question asks for the $2^{	ext{nd}}$ harmonic of the pipe in its new state. The $2^{	ext{nd}}$ harmonic (first overtone) of a closed pipe is $3f'_1$.Therefore,the frequency of the $2^{	ext{nd}}$ harmonic is $3 	imes 600 \,Hz = 1800 \,Hz$.

The fundamental frequency of a closed pipe is $400 \,Hz$. If $1/3$ of the pipe is filled with water,the frequency of the $2^{\text{nd}}$ harmonic of the pipe will be (Neglect end correction). (in $\,Hz$)

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