The angular momentum of a solid cylinder rotating about its geometric axis with angular speed $40rad{s}^{-1}is2kg-m^2{s}^{-1}.$ If the radius of the cylinder is 10cm, the mass of the cylinder is

 
For a solid cylinder rotating about it’s axis angular momentum L is given by
 $L=I\omega \mathrm{....}\left(i\right)$
Where, I = moment of inertia$=\frac{1}{2}M{R}^2$
Substituting $L=2kg-m^2{s}^{-1}, \omega =40rad{s}^{-1}, R=10\times {10}^{-2}m$
We get,  $2=\frac{1}{2}\left(M\right){(10\times {10}^{-2})}^2\left(40\right)\Rightarrow M=10kg$