JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Waves & Sound

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 He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas).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
(Useful information:

167 R T = 640 J^{1/2}

mole

^{- 1/2}; 140 R T = 590 J^{1/2}

mole

^{- 1/2}

.The molar masses

M

in grams are given in the options. Take the values of

\frac{10}{M}

for each gas as given there.)

(A)Neon $(M = 20, \frac{10}{20} = \frac{7}{10})$
(B)Nitrogen $(M = 28, \frac{10}{28} = \frac{3}{5})$
(C) $Oxygen (M = 32, \frac{10}{32} = \frac{9}{16})$
(D) $Argon (M = 36, \frac{10}{36} = \frac{17}{32})$ ✓ Correct

Answer: (D)

Solution

At first resonance,

\frac{λ}{4} = 0.350 \pm 0.005

m, so

λ = 1.400 \pm 0.020

m.

Wave speed

v = f λ = 244 \times (1.400 \pm 0.020) = (341.6 \pm 4.88)

m/s.

For an ideal gas,

v = \frac{γ R T}{M} = \frac{100 γ R T}{100 M} = \frac{100 γ R T}{100 M \times 1 0 ^{- 3}} = \frac{100 γ R T}{0.1 M} = \frac{1000 γ R T}{M}

.

For monoatomic gas (

γ = 5/3

167 R T = 640

, so

v = 640 \frac{10}{M}

.
For diatomic gas (

γ = 7/5

140 R T = 590

, so

v = 590 \frac{10}{M}

.

Checking options:
Neon (monoatomic):

640 \times \frac{7}{10} = 448

m/s
Nitrogen (diatomic):

590 \times \frac{3}{5} = 354

m/s
Oxygen (diatomic):

590 \times \frac{9}{16} = 332

m/s
Argon (monoatomic):

640 \times \frac{17}{32} = 340

m/s

Argon's value of 340 m/s falls within the range

341.6 \pm 4.88

m/s (336.7 to 346.5 m/s).

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Waves & Sound

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