On producing the waves of frequency 1000 ,Hz | vedclass

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(B) Frequency, $f = 1000 \,Hz$.Total distance between $6$ successive nodes is $d = 85 \,cm = 0.85 \,m$.In a stationary wave, the distance between two

Vedclass · Accepted Answer

$340$

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(B) Frequency, $f = 1000 \,Hz$.Total distance between $6$ successive nodes is $d = 85 \,cm = 0.85 \,m$.In a stationary wave, the distance between two successive nodes is $\frac{\lambda}{2}$.Therefore, the distance between $6$ successive nodes (which involves $5$ intervals of $\frac{\lambda}{2}$) is given by:$d = 5 	imes \frac{\lambda}{2} = \frac{5 \lambda}{2}$.Equating this to the given distance:$\frac{5 \lambda}{2} = 0.85 \,m$.$\lambda = \frac{0.85 	imes 2}{5} = 0.34 \,m$.The speed of sound $v$ is given by $v = f \lambda$.$v = 1000 \,Hz 	imes 0.34 \,m = 340 \,ms^{-1}$.

On producing the waves of frequency $1000 \,Hz$ in a Kundt's tube, the total distance between $6$ successive nodes is $85 \,cm$. Then, the speed of sound in the gas filled in the tube is (in $\,ms^{-1}$)

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