A disc rotates about its axis of symmetry in a hoizontal plane at a...

Using, $\mu m g=\frac{m v^2}{r}=m r \omega^2\omega=2 \pi n=2 \pi \times 3.5=7 \pi \mathrm{rad} / \mathrm{sec}$ Radius, $r=1.25 \mathrm{~cm}=1.25 \times 10^{-2} \mathrm{~m}$ Coefficient of friction, $\mu=$ ? \mu m g=$\frac{m(r \omega)^2}{r}(\because$v=r $\omega)<img src="/api/proxy-image?url=https$ \\ \begin{gathered} \Rightarrow \quad \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10}\\ =\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6 \end{gathered}