A solid sphere of radius $R$ starts rotating on rough horizontal surface with translational velocity $V_0$ and initial angular velocity ${\omega }_0=\frac{2V_0}{3R}$ . The sphere starts pure rolling after some time $t$. if $V$ is the translational velocity at pure rolling. Assume uniformly accelerated motion up to start of pure rolling:
if $S$ is the distance covered by the sphere in rotational motion up to the instant at which pure rolling starts, find the velocity (translational) of the sphere during the pure rolling

The correct option is C $\frac{38}{33}\frac{S}{t}$
Draw a diagram of given problem.

Find angular acceleration of the sphere.
Formula used:$\omega ={\omega }_0+\alpha t$
Given,
${\omega }_0=\frac{2V_0}{3R}$

$V_0=\frac{3}{2}R{\omega }_0$
according to diagram, angular momentum is conserved about $A,$
$L_1=L_2$
Therefore,
$\frac{2}{5}m{R}^2{\omega }_0+m{V}_{0}R=\frac{2}{5}m{R}^2\omega +mVR$

$\frac{2}{5}m{R}^2\times \frac{2V_0}{3R}+m{V}_{0}R=\frac{2}{5}m{R}^2\frac{V}{R}+mVR$

$V_0=\frac{21}{19}V\left(i\right)$
Now, according to question $\omega$ and $V$ are translational velocity for pure rolling starts,
$\omega ={\omega }_0+\alpha t$
From equation $\left(i\right)$,we get
$\alpha =\frac{5V}{19Rt}$
Where, $t$ is time when pure rolling starts.

Find angle by which sphere rotates.
Formula used:$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
Pure rolling starts when sphere start rotated by an angle $\theta$,
Therefore,
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$\theta =\frac{2V_0}{3R}t+\frac{1}{2}.\frac{5V}{19Rt}.t^2$

$\theta =\frac{33}{38}\frac{Vt}{R}$

$R\theta =S=\frac{33}{38}Vt$
Therefore,
Translational velocity$\left(V\right),$
$V=\frac{38}{33}\frac{S}{t}$

Final answer: (d)