On producing the waves of frequency \(1000 \mathrm{~Hz}\) in a Kundt's tube, the total distance between 6…
On producing the waves of frequency \(1000 \mathrm{~Hz}\) in a Kundt's tube, the total distance between 6 successive nodes is \(85 \mathrm{~cm}\). Then, the speed of sound in the gas filled in the tube is
\(330 \mathrm{~ms}^{-1}\)
\(340 \mathrm{~ms}^{-1}\)
\(350 \mathrm{~ms}^{-1}\)
\(300 \mathrm{~ms}^{-1}\)
Solution
Frequency, \(f=1000 \mathrm{~Hz}\)
Total distance between 6 successive nodes
\(d=85 \mathrm{~cm}=0.85 \mathrm{~m}\)
Since, distance between 6 successive nodes is equivalent to \(2 \lambda+\frac{\lambda}{2}=\frac{5 \lambda}{2}\)
\(\therefore \quad d=\frac{5 \lambda}{2}=0.85 \Rightarrow \lambda=\frac{0.85 \times 2}{5}=0.34 \mathrm{~m}\)
\(\therefore\) Speed of sound in tube,
\(v=f \lambda=1000 \times 0.34=340 \mathrm{~ms}^{-1}\)
