The frequency of the third overtone of a pipe of length ' $L_c$ ', closed at one end is same as the…

The frequency of the third overtone of a pipe of length ' $L_c$ ', closed at one end is same as the frequency of the sixth overtone of a pipe of length ' $L_0$ ', open at both ends. Then the ratio $\mathrm{L}_{\mathrm{c}}: \mathrm{L}_0$ is
  1. $1: 4$
  2. $1: 2$
  3. $2: 1$
  4. $4: 1$

Solution

Third overtone of pipe closed at one end $\mathrm{f}=\frac{(2 \mathrm{n}+1) \mathrm{V}}{4 l_1}=\frac{(2 \times 3+1) \mathrm{V}}{4 l_1}=\frac{7 \mathrm{~V}}{4 l_1}$
Sixth overtone of pipe open at both ends $\begin{aligned} & \mathrm{f}=\frac{\mathrm{nV}}{2 l_2}=\frac{7 \mathrm{~V}}{2 l_2} \Rightarrow \frac{7 \mathrm{~V}}{4 l_1}=\frac{7 \mathrm{~V}}{2 l_2} \\ & \frac{\mathrm{~L}_{\mathrm{c}}}{\mathrm{~L}_{\mathrm{o}}}=\frac{1}{2} \quad \ldots\left(\because \mathrm{l}_1=\mathrm{L}_{\mathrm{C}} \text { and } \mathrm{l}_2=\mathrm{L}_{\mathrm{O}}\right) \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 2)

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