MHT CET · Physics · Waves and Sound
The end correction of resonance tube is 1 cm . If the shortest length resonating with a tuning fork is 15 cm , the next resonating length will be
- A 35 cm
- B 40 cm
- C 47 cm
- D 64 cm
Answer & Solution
Correct Answer
(C) 47 cm
Step-by-step Solution
Detailed explanation
Let the shortest length be denoted as
\(l_1=15.0 \mathrm{~cm}\)
The end correction, \(\mathrm{e}=1.0 \mathrm{~cm}\)
\(l_1+\mathrm{e}=\frac{\lambda}{4}\)
Now, \(l_2+\mathrm{e}=\frac{3 \lambda}{4}\)
\(\begin{array}{ll}
\therefore & \frac{l_1+\mathrm{e}}{l_2+\mathrm{e}}=\frac{1}{3} \\
\therefore & \frac{15+1}{l_2+1}=\frac{1}{3} \\
\therefore & l_2+1=48 \\
\therefore & l_2=47 \mathrm{~cm}
\end{array}\)
\(l_1=15.0 \mathrm{~cm}\)
The end correction, \(\mathrm{e}=1.0 \mathrm{~cm}\)
\(l_1+\mathrm{e}=\frac{\lambda}{4}\)
Now, \(l_2+\mathrm{e}=\frac{3 \lambda}{4}\)
\(\begin{array}{ll}
\therefore & \frac{l_1+\mathrm{e}}{l_2+\mathrm{e}}=\frac{1}{3} \\
\therefore & \frac{15+1}{l_2+1}=\frac{1}{3} \\
\therefore & l_2+1=48 \\
\therefore & l_2=47 \mathrm{~cm}
\end{array}\)
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