A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision. [Ignore friction impulse by ground during collision]

The data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact. From linear momentum conservation in horizontal direction, we have
 $(-2\times 1)+(0.1\times 0)+(2\times v)$     
Here, v is the velocity of CM of ring impact. Solving the above equation, we have v=0
Thus, CM becomes stationary 
$\therefore$ Correct option is (a) 
Linear impulse during impact 
(i) in horizontal direction
$J_1=\Delta p=0.1\times 20=2N-s$ 
(ii) In vertical direction
$J_2=\Delta p=0.1\times 20=1N-s$ 
Writing the equation (about CM) 

Angular impulse=change in angular momentum
We have 
$1\times (\frac{\sqrt{3}}{2}\times \frac{1}{2})-2\times 0.5\times \frac{1}{2}=2\times {\left(0.5\right)}^2[\omega -\frac{1}{0.5}]$ 
Solving this question $\omega$  comes out be positive or  $\omega$ antic-clockwise. So just after collision rightwards slipping is taking place.
Here, friction is leftwards.
Therefore, option c is also correct.