A student is performing an experiment using a resonance column and 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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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(es) in the tube is (Useful information: $\sqrt{167 R T} = 640 J^{1 / 2} m o l e^{- 1 / 2}$ ; $\sqrt{140 R T} = 590 J^{1 / 2} m o l e^{- 1 / 2}$ . The molar masses $M$ in grams are given in the options. Take the values of $\sqrt{\frac{10}{M}}$ for each gas as given there.)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$A r g o n (M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32})$

$O x y g e n (M = 32, \sqrt{\frac{10}{32}} = \frac{9}{16})$

$N i t r o g e n (M = 28, \sqrt{\frac{10}{28}} = \frac{3}{5})$

$N e o n (M = 20, \sqrt{\frac{10}{20}} = \frac{7}{10})$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here $υ = \frac{v}{4 l} = \sqrt{\frac{γ R T}{M \times 10^{- 3}}} \times \frac{1}{4 l} \Rightarrow v = υ λ = υ \times 4 l$
$\Rightarrow v (244) \times 4 \times l = 336.7 m / s to 346.5 m / s [∴ l = 0.350 \pm 0.005]$
For monatomic gas $γ = 1.67$
$∴ v = \sqrt{\frac{γ R T}{M \times 10^{- 3}}} = \sqrt{100 γ R T} \times \sqrt{\frac{10}{M}} = \sqrt{167 R T} \times \sqrt{\frac{10}{M}} = 640 \sqrt{\frac{10}{M}}$
For Neon $M = 20 ∴ v = 640 \times \frac{7}{10} = 448 m s^{- 1}$
For argon $M = 36, ∴ v = 640 \times \frac{17}{32} = 340 m s^{- 1}$
For diatomic gas $γ = 1.4$
$v = \sqrt{140 R T} \sqrt{\frac{10}{M}} = 590 \times \sqrt{\frac{10}{M}}$
For Oxygen $M = 32 ∴ v = 590 \times \frac{9}{16} = 331.87 m s^{- 1}$
For Nitrogen $M = 28 ∴ v = 590 \times \frac{3}{5} = 354 m s^{- 1}$