A disc of mass 100g and radius 10cm has a projection on its circumference. The mass of projection is negligible. A 20g bit of putty moving tangential to the disc with a velocity of 5ms^-1 strikes the projection and sticks to it. The angular velocity of disc is

In this instance, the angular momentum of the putty bit with respect to the axis of rotation equals the angular momentum of the disc and putty bit system with respect to the axis of rotation.
Let:

Mass of putty is M = 20g

m = the disc's mass = 100g

$$\begin{gathered} \mathrm{MvR}=\left(\frac{{\mathrm{mR}}^2}{2}+{\mathrm{MR}}^2\right)\mathrm{\omega } \\ \mathrm{\omega }=\frac{\mathrm{MvR}}{\left(\frac{{\mathrm{mR}}^2}{2}+{\mathrm{MR}}^2\right)} \\ \mathrm{\omega }=\frac{20\times 500\times 10}{{\displaystyle \frac{100\times {10}^2}{2}+20\times {10}^2}} \\ \mathrm{\omega }=14.29 \mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence the correct answer is 14.29 rad/s.