Question

Two molecules of gas have speeds of `9 xx 10^(6) ms^(-1)` and `1 xx 10^(6) ms^(-1)` respectively. What is the root mean square speed of these molecules?
A. `(sqrt(21)) xx 10^(6) m//s`
B. `(sqrt(41)) xx 10^(6) m//s`
C. `8.5 xx 10^(6) m//s`
D. `(sqrt(17)) xx 10^(6) m//s`

Answer 1

Correct Answer - B
For `n` - molecules we know that
`v_(rms) = sqrt((v_(1)^(2)+v_(2)^(2)+v_(3)^(2)+...+v_(n)^(2))/(n))`
`[v_(rms)`= root mean square velocity]
where, `v_(1) , v_(2), v_(3) ....v_(n)` are individual velocities of `n`- molecules of the gas.
For two molecules, `v_(rms) = sqrt((v_(1)^(2)+v_(2)^(2))/(2))`
Given `v_(1) = 9xx10^(6)m//s` and `v_(2) = 1 xx 10^(6)m//s`
`:. v_(rms) = sqrt((9xx10^(6))^(2)+(1xx10^6)^(2))/(2) = sqrt(41) xx 10^(6)m//s`.