A student is performing an experiment using a resonance column and a tuning fork of frequency `244s^(-1)`. He is told that the air in the tube has been replaced by another gas (assuming that the air column remains filled with the gas). If the minimum height at which resonance occurs is `(0.350+- 0.005)m`, the gas in the tube is (Useful information : `sqrt(167RT) = 640J^(1//2)mol^(-1//2)`,
`sqrt(140RT) = 590J^(1//2)mol^(-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.)

Options

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))`

Answer

D

Explanation

`V=sqrt((gammaRT)/(m))` [m should be taken in Kg]
Neon and argon are monoatomic gas `implies gamma= 1.67`
Nitrogen and oxygen are diatomic gas `implies gamma = 1.4`
`therefore V = sqrt((1.67 xx RT xx 1000)/(m))` (m is in grams)
`therefore` For Neon `V= sqrt(167RTxx (10)/(20)) = 640 xx (7)/(10)m//s = 448m//s`
For Nitrogen , `V= sqrt(140RT xx (10)/(28)) = 590 xx(3)/(5)m//s = 354m//s`
For oxygen, `V= sqrt(140RT xx (10)/(32)) = 590xx (9)/(16) m//s = 331.875m//s`
From the given information , f =` 244` Hz
`(gamma)/(4)|_(max) = 0.350+0.005 = 0.355m`
`(gamma)/(4)|_(max) = 0.350-0.005 = 0.345m`
`therefore lambda_(max) = 1.420m`
`lambda_(min) = 1.380m`
`therefore V_(max) = 244 xx lambda_(max) = 346.48 "m"//"sec"`
`V_("min") = 244 xx lambda_("min") = 336.72 "m"//"sec"`
Only Argon is between them