JEE Advanced 2014 · Paper 1 · Physics Q6

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PhysicsMultiple correct+3Question 6 of 60

A student is performing an experiment using a resonance column and a tuning fork of frequency $244\ \mathrm s^{-1}$ . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350\pm0.005)$ m, the gas in the tube is

(Useful information: $\sqrt{167RT}=640\ \mathrm{J^{1/2}\,mole^{-1/2}}$ ; $\sqrt{140RT}=590\ \mathrm{J^{1/2}\,mole^{-1/2}}$ . The molar masses $M$ in grams are given in the options. Take the values of $\sqrt{10/M}$ for each gas as given there.)

Neon

\left(M=20,\ \sqrt{\tfrac{10}{20}}=\tfrac7{10}\right)

Nitrogen

\left(M=28,\ \sqrt{\tfrac{10}{28}}=\tfrac35\right)

Oxygen

\left(M=32,\ \sqrt{\tfrac{10}{32}}=\tfrac9{16}\right)

Argon

\left(M=36,\ \sqrt{\tfrac{10}{36}}=\tfrac{17}{32}\right)

Answer and worked solution

Official answer: D

Speed of sound from the resonance. The first resonance is at $L=\lambda/4$ (end correction neglected), so
$v=4fL=4\times244\times0.350=341.6\ \mathrm{m\,s^{-1}},$
and the range $0.345$ - $0.355$ m gives $336.7$ - $346.5\ \mathrm{m\,s^{-1}}$ .

Each gas. $v=\sqrt{\dfrac{\gamma RT}{M}}$ with $M$ in kg $=M_g/1000$ :
$v=\sqrt{\gamma RT\cdot100}\,\sqrt{\frac{10}{M_g}}.$
Monatomic gases ( $\gamma=1.67$ ): $v=640\sqrt{10/M_g}$ . Diatomic ( $\gamma=1.4$ ): $v=590\sqrt{10/M_g}$ .

Gas	$v$ (m s⁻¹)
Neon	$640\times0.7=448$
Nitrogen	$590\times\frac35=354$
Oxygen	$590\times\frac9{16}=332$
Argon	$640\times\frac{17}{32}=340$

Only argon falls within the range.

(D)

Topics:Organ pipes and resonance Speed of waves Speed of sound by resonance tube

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