
- ✓$\frac{1}{2}$
- B$1$
- C$\frac{3}{2}$
- D$2$

So we can write $f=m a$ So $a=\frac{f}{m}$
$f \cdot r=I \alpha$
For the disc $I=\frac{1}{2} m r^{2}$
$\alpha=\frac{2 f}{m r}$
If the disc stops moving after travelling for a while, both $v_{0}$ and $\omega_{0}$ become zero at the same time.
If time $t$ required to stop, we can write
$v_{0}-a t=0$ and $\omega_{0}-\alpha t=0$
$\therefore \frac{v_{0}}{a}=\frac{\omega_{0}}{\alpha}$
$\therefore \frac{v_{0} m}{f}=\frac{\omega_{0} r m}{2 f}$
$\therefore \frac{v_{0}}{\omega_{0} r}=\frac{1}{2}$
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[Given: Surface tension of the liquid is $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]
