A) \[{{L}_{0}}=\frac{L}{2},{{L}_{C}}=2L\]
B) \[{{L}_{0}}=\frac{L}{4},{{L}_{C}}=2L\]
C) \[{{L}_{0}}=\frac{L}{2},{{L}_{C}}=4L\]
D) \[{{L}_{0}}=\frac{L}{4},{{L}_{C}}=4L\]
Correct Answer: A
Solution :
\[{{f}_{0}}-{{f}_{c}}=2\] \[\frac{V}{21}\,-\frac{V}{41}\,=2\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\frac{V}{l}\,=8\] When lengths of both pipes are changed \[f_{0}^{'}\,-f_{c}^{'}\,=7\] \[=\left( \frac{7}{8}\, \right)\times \,8=\frac{7}{8}\,\times \left( \frac{V}{1} \right)\] \[\left( \sin ce\,\,\frac{V}{l}=8 \right)\] \[\frac{V}{{{n}_{1}}l}\,-\frac{V}{{{n}_{2}}l}\,=\frac{7}{8}\,\frac{V}{l}\] It may be possible for \[{{n}_{1}}=1\] and\[{{n}_{2}}=8\] so that \[\frac{V}{l}-\frac{V}{8l}\,=\frac{7}{8}\frac{V}{l}\] \[\Rightarrow \,\frac{V}{2\left( \frac{L}{2} \right)}\,-\frac{V}{4(2L)}\,=\frac{7}{8}\,\frac{V}{L}\]So length of open pipe is halved and length of closed pipe is doubled.You need to login to perform this action.
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