Correct Answer - (a) `approx 409.9 m//s` , (b) `(1)/(6) %`
(a) `gamma_(mix) = (n_(1)C_(P_(1)) + n_(2)C_(P_(2)))/ (n_(1)C_(V_(1)) + n_(2)C_(V_(2))) = (1xx(5)/(2)R+2xx(7)/(2)R)/(1xx(3)/(2)R+2xx(5)/(2)R) = (19)/(13)`
`m_(mix) = (n_(1)m_(1) + n_(2)m_(2))/(n_(1) + n_(2)) = (1 xx 4 + 2 xx 32)/(1 + 2) = (68)/(3)`
`V = sqrt((gamma_(mix)RT)/(m_(mix))) = sqrt((19 xx 25 xx 300 xx 3)/(13 xx 3 xx 68 xx 10^(-1))) = 400.9 m//sec`
(b) `V = sqrt((gammaRT)/(M)) rArr ln V = (1)/(2)ln"(gammaR)/(M) + (1)/(2) ln T`
`(1)/(V) (dV)/(dT) = 0 + (1)/(2T)`
`(dV)/(V) xx 100 = (1)/(2) (dT)/(T) xx 100 = (1)/(2 xx 300) xx 100 = (1)/(6)%`