A sample of oxygen at `NTP` has volume `V` and a sample of hydrogen at `NTP` has volume `4 V`. Both the gases are mixed and the mixture is maintained at `NTP` if the speed of sound in hydrogen at `NTP` is `1270 m//s`, that in the mixture will be

Options

A. `317 m//s`
B. `635 m//s`
C. `830 m//s`
D. `950 m//s`

Answer

B

Explanation

`rho_(max) = (rho_(O_(2)) V_(O_(2)) + rho_(H_(2)) V_(N_(2)))/(V_(O_(2)) + V_(H_(2)))`
`= (16 xx V + 1 xx 4V)/(V + 4V)`
`(v_("mix"))/(v_(H_(2))) = sqrt((rho_(H_(2)))/(rho_("mix"))) = sqrt((1)/(4)) = (1)/(2)`
`v_("mix") = (v_(H_(2)))/(2) = (1270)/(2) = 635 m//s`