Wave Motion 5 Question 15
15. A student is performing an experiment using a resonance column and a tuning fork of frequency $244 s^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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 $(0.350 \pm 0.005) m$, the gas in the tube is
(2014 Adv.) (Useful information : $\sqrt{167 R T}=640 J^{1 / 2} mole^{-1 / 2}$; $\sqrt{140 R T}=590 J^{1 / 2} mole^{-1 / 2}$. The molar masses $M$ in grams are given in the options. Take the value of $\sqrt{10 / M}$ for each gas as given there.)
(a) Neon $(M=20, \sqrt{10 / 20}=7 / 10)$
(b) Nitrogen $(M=28, \sqrt{10 / 28}=3 / 5)$
(c) Oxygen $(M=32, \sqrt{10 / 32}=9 / 16)$
(d) Argon $(M=36, \sqrt{10 / 36}=17 / 32)$
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Solution:
- Minimum length $=\frac{\lambda}{4}$
$$ \lambda=4 l $$
Now,
$$ v=f \lambda=(244) \times 4 \times l $$
as
$$ l=0.350 \pm 0.005 $$
$\Rightarrow v$ lies between $336.7 m / s$ to $346.5 m / s$
Now, $v=\sqrt{\frac{\gamma R T}{M \times 10^{-3}}}$, here $M$ is molecular mass in gram
$$ =\sqrt{100 \gamma R T} \times \sqrt{\frac{10}{M}} $$
For monoatomic gas, $\gamma=1.67$
$$ \Rightarrow \quad v=640 \times \sqrt{\frac{10}{M}} $$
For diatomic gas, $\gamma=1.4$
$\Rightarrow v=590 \times \sqrt{\frac{10}{M}} \Rightarrow v _{Ne}=640 \times \frac{7}{10}=448 m / s$ $\Rightarrow v _{Ar}=640 \times \frac{17}{32}=340 m / s \Rightarrow v _{O _2}=590 \times \frac{9}{16}=331.8 m / s$
$\Rightarrow v _{N _2}=590 \times \frac{3}{5}=354 m / s$
$\therefore$ Only possible answer is Argon.
