A pipe open at both ends of length 1.5 ,m is | vedclass

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(B) Let $L = 1.5 \,m$ be the total length of the pipe and $\ell$ be the length immersed in water. The length of the air column above the water is $L'

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) Let $L = 1.5 \,m$ be the total length of the pipe and $\ell$ be the length immersed in water. The length of the air column above the water is $L' = L - \ell = 1.5 - \ell$.Since the pipe is now closed at one end (by water), it acts as a closed organ pipe.The frequency of the $n^{th}$ overtone for a closed pipe is given by $f_n = \frac{(2n+1)v}{4L'}$, where $n$ is the overtone number.For the second overtone, $n = 2$, so the frequency is $f_2 = \frac{(2(2)+1)v}{4L'} = \frac{5v}{4L'}$.Given $f_2 = 330 \,Hz$ and $v = 330 \,m/s$, we have:$330 = \frac{5 	imes 330}{4(1.5 - \ell)}$$1 = \frac{5}{4(1.5 - \ell)}$$4(1.5 - \ell) = 5$$6 - 4\ell = 5$$4\ell = 1$$\ell = 0.25 \,m$.

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