A pipe of length 1.5 m closed at one end is f | vedclass

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(B) For an open pipe,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.Given $v = 360\ m/s$ and $l = 1.5\ m$,we have $n_0 = \frac{360}{2 	im

Vedclass · Accepted Answer

$683$

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(B) For an open pipe,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.Given $v = 360\ m/s$ and $l = 1.5\ m$,we have $n_0 = \frac{360}{2 	imes 1.5} = \frac{360}{3} = 120\ Hz$.For a closed pipe,the fundamental frequency is $n_1 = \frac{v'}{4l}$,where $v'$ is the speed of sound in the gas at $30\,^{\circ}C$.Since both pipes resonate with the same tuning fork,$n_1 = n_0 = 120\ Hz$.Thus,$120 = \frac{v'}{4 	imes 1.5} \Rightarrow 120 = \frac{v'}{6} \Rightarrow v' = 720\ m/s$.Using the relation $v \propto \sqrt{T}$,the speed of sound in the gas at $0\,^{\circ}C$ $(v_0)$ is related to the speed at $30\,^{\circ}C$ $(v')$ by $v_0 = v' 	imes \sqrt{\frac{T_0}{T_{30}}}$.Here $T_0 = 273\ K$ and $T_{30} = 273 + 30 = 303\ K$.$v_0 = 720 	imes \sqrt{\frac{273}{303}} = 720 	imes \sqrt{0.90099} \approx 720 	imes 0.9492 \approx 683.4\ m/s$.Rounding to the nearest integer,the speed is $683\ m/s$.

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