The wavelengths of two notes in air are frac{ | vedclass

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(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$. The frequency of a note is given by $f = \frac{v}{\lambda}$.For the f

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$360$

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(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$. The frequency of a note is given by $f = \frac{v}{\lambda}$.For the first note,frequency $f_1 = \frac{v}{36/195} = \frac{195v}{36}$.For the second note,frequency $f_2 = \frac{v}{36/193} = \frac{193v}{36}$.Since both produce $10$ beats per second with the third note,we have:$\frac{195v}{36} - n = 10$  ...$(i)$$n - \frac{193v}{36} = 10$  ...(ii)Adding equations $(i)$ and (ii):$\frac{195v}{36} - \frac{193v}{36} = 20$$\frac{2v}{36} = 20$$\frac{v}{18} = 20$$v = 360 ~m/s$.

The wavelengths of two notes in air are $\frac{36}{195} ~m$ and $\frac{36}{193} ~m$. Each note produces $10$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

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