Question

In Quincke’s tube a detector detects minimum intensity. Now one of the tubes is displaced by 5 cm. During displacement the detector detects maximum intensity 10 times, then finally a minimum intensity (When displacement is complete). The wavelength of sound is:
A. $\dfrac{10}{9}cm$
B. $1cm$
C. $\dfrac{1}{2}cm$
D. $\dfrac{5}{9}cm$

Answer 1

Hint: A Quincke’s tube is a device which is used to demonstrate the interference effects in standing sound waves. This device consists of a resonance tube with a millimetre scale, is partially filled with water and is connected to an expansion vessel with a tube.

Formula used:
For solving the given question, we will be using the formula for maximum intensity sound changes
\[\Delta P=n\dfrac{\lambda }{2}\]

Complete step by step answer:
 Before solving the question, let us take a look the given parameters
Since 10 maxima points were found while moving the tube by 5 cm, i.e. the displaced by 5 cm, we can safely say that the distance is equal to the 5 waves or we can say 10 half waves
So, we have
\[\Delta P=5cm\]
n = 10
$\lambda =wavelength$
Therefore, we can see that
\[\Rightarrow \Delta P=n\times \dfrac{\lambda }{2}\]
\[\Rightarrow 5=10\times \dfrac{\lambda }{2}\]
\[\Rightarrow \dfrac{5}{10}=\dfrac{\lambda }{2}\]
\[\lambda =1cm\]

Note:
The Quincke’s tube is named after the German physicist Georg Hermann Quincke. He invented the Quincke’s Tube in the 19th century while working on the phenomenon of interference of sound waves. This device works on two major points
The intensity of sound changes to minimum when displaced by distance, \[\Delta P=(2n-1)\dfrac{\lambda }{2}\]
The intensity of sound changes to minimum when displaced by distance, \[\Delta P=n\dfrac{\lambda }{2}\]
For this question we have used the formula for the case of minimum intensity.