A hypothetical gas sample has its molecular speed distribution graph as shown in  the figure. Speed u and $\frac{dN}{du}$  have appropriate limits. The root mean square speed  of the molecules is $\sqrt{\frac{N}{3}}$. Find N. (Do not worry about units).

By definition,  $v_{RMS}=\sqrt{\frac{{\displaystyle \underset{0}{\overset{N_0}{\int }}{u}^{2}dN}}{N_0}}$, where  $N_0$ is total number of molecules.
Equation of line is:  $\frac{dN}{du}=4-u$
$\Rightarrow dN=\left(4-u\right)du$ 
$\Rightarrow v_{RMS}=\sqrt{\frac{{\displaystyle \underset{0}{\overset{4}{\int }}{u}^2\left(4-u\right)du}}{N_0}}$  (Here integration only runs from 0 to 4 because max value  of u is 4.
Also, area under the graph =  $\int \frac{dN}{du}du=\frac{1}{2}\left(4\right)\left(4\right)=8=N_0$
Hence,  $v_{RMS}=\sqrt{\frac{{\displaystyle \underset{0}{\overset{4}{\int }}{u}^2\left(4-u\right)du}}{8}}=\sqrt{\frac{1}{8}\left(\frac{256}{3}-64\right)}=\sqrt{\frac{8}{3}}$.