MHT CET · Physics · Waves and Sound
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
- A \(1: 4\)
- B \(1: 2\)
- C \(2: 1\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(B) \(1: 2\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
& \mathrm{f}=\frac{\mathrm{nV}}{2 l_2}=\frac{7 \mathrm{~V}}{2 l_2} \Rightarrow \frac{7 \mathrm{~V}}{4 l_1}=\frac{7 \mathrm{~V}}{2 l_2} \\
& \frac{\mathrm{~L}_{\mathrm{c}}}{\mathrm{~L}_{\mathrm{o}}}=\frac{1}{2} \quad \ldots\left(\because \mathrm{l}_1=\mathrm{L}_{\mathrm{C}} \text { and } \mathrm{l}_2=\mathrm{L}_{\mathrm{O}}\right)
\end{aligned}\)
\(\mathrm{f}=\frac{(2 \mathrm{n}+1) \mathrm{V}}{4 l_1}=\frac{(2 \times 3+1) \mathrm{V}}{4 l_1}=\frac{7 \mathrm{~V}}{4 l_1}\)
Sixth overtone of pipe open at both ends
\(\begin{aligned}
& \mathrm{f}=\frac{\mathrm{nV}}{2 l_2}=\frac{7 \mathrm{~V}}{2 l_2} \Rightarrow \frac{7 \mathrm{~V}}{4 l_1}=\frac{7 \mathrm{~V}}{2 l_2} \\
& \frac{\mathrm{~L}_{\mathrm{c}}}{\mathrm{~L}_{\mathrm{o}}}=\frac{1}{2} \quad \ldots\left(\because \mathrm{l}_1=\mathrm{L}_{\mathrm{C}} \text { and } \mathrm{l}_2=\mathrm{L}_{\mathrm{O}}\right)
\end{aligned}\)
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