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

Explanation

If `V_(H)` and `V_(m)` are the velocities in hydrogen and mixture respectively, then
`(v_(m))/(v_(H))=sqrt((rho_(H))/(rho_(m)))`……….i
Density of mixture `rho_(m)=(rho_(0)V_(0)+rho_(H)V_(H))/(V_(o)+V_(H))`
where `rho_(0)` and `V_(0)` are the density of volume of the oxygen.
`rho_(m)=(rho_(H)v_(H)(1+(rho_(0))/(rho_(H))xx(V_(0))/(V_(H))))/(V_(H)(1+(V_(0))/(V_(H))))`
or `(rho_(m))/(rho_(H))=((1+(rho_(0))/(rho_(H))xx(V_(0))/(v_(H))))/((1+(V_(0))/(V_(H))))`
`=(1+16xx1/4)/(1+1/4)=4`
or `(rho_(H))/(rho_(k))=1/4`
From equation (i)
`v_(m)=v_(H)sqrt(1/4)=(v_(H))/2`
`=1270/2=635m//s`