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(D) The tube is closed at one end, so it acts as a closed organ pipe. The lowest frequency (fundamental frequency) produced in the tube corresponds to

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$1.4$

Vedclass · Answer

(D) The tube is closed at one end, so it acts as a closed organ pipe. The lowest frequency (fundamental frequency) produced in the tube corresponds to the condition where the length of the tube $l = \frac{\lambda}{4}$.Given: $l = 1.0 	ext{ m}$, $f = 84 	ext{ Hz}$.Therefore, the wavelength $\lambda = 4l = 4 	imes 1.0 = 4 	ext{ m}$.The velocity of sound in air is given by $v = f \lambda = 84 	imes 4 = 336 	ext{ m/s}$.The velocity of sound is also given by the formula $v = \sqrt{\frac{\gamma P}{ho}}$, where $P$ is the atmospheric pressure and $ho$ is the density of air.Squaring both sides, we get $v^{2} = \frac{\gamma P}{ho}$.Rearranging for $\gamma$, we get $\gamma = \frac{v^{2} ho}{P}$.Substituting the values: $\gamma = \frac{(336)^{2} 	imes 1.2}{1.0 	imes 10^{5}} = \frac{112896 	imes 1.2}{100000} = \frac{135475.2}{100000} \approx 1.355$.Rounding to the nearest option, we get $\gamma = 1.4$.

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