Just one question why we have converted Omega here in rad/s we can also apply formula in rev/s

no reason i always just convert it for getting all in standard terms

Is it D?

Assume that sphere is moving towards right with a speed v and angular velocity \omega (rotating in clock-wise direction).

After collision, v is towards left. The point in contact with the ground is not at rest now (its speed is v+r\omega). So, it will be slipping until the sphere acquires rolling without slipping, which will happen only after some time.

So D is the correct answer.

Why is it necessary that the sphere will have angular velocity omega a long time after the collision? @Cooljet123

Friction continues to act until no slipping condition is achieved (I meant any general omega, not the initial one)…let me change it. Thanks for pointing out.

Is there any condition to apply

Torque = I × alpha

Means when we can’t apply it

torque =I * alpha is applicable only about COM or abt points which r at rest if u want to apply from a moving point then we need to consider pseudo force passing throught COM also

Can u send such a question where we need to make use of pseudo torque? (If u have such a question). Yeah like a question where we are forced to calculate pseudo torque

Do you mean where pseudo force produces a Torque? Or something else?

Bro always apply torque at COM that makes the torque of pseudo force zero so it’s quite helpful to us.

Well that means it do not have a condition right ?

no it has conditions

the torque equation tou=I*alpha is used only abt a point which belongs to the body only if it doesnt belong to the body then u cant apply that equation
i ll give an example
case 1:let us assume that a rod is rotating abt an axis passing through COM on a horizontal plane now if u want to apply torque abt an end point of rod then u ll write pseudo force torque and torque due to axis
case 2: try doing same abt a point on the ground just below the point u considerd above as there is no pseudo force (its belonging to ground hece it ll not move) now torque is only due to axis of rotation hence u ll get different angular acclerations
but do u think u ll observe 2 different acclerations in both casses ans is no
in second case u cannot apply that equation u need to differentate angular momentum
L=m r×vcom+icom*omega

differentaite above one u ll get same angular accleration as in case 1

My 3 equations : linear momentum conservation , Angular momentum conservation , coefficient of restitution but still not getting the answer .

Is it single correct ?

I’m getting W = V/L

omega about the rod’s centre

Didn’t check that energy terms…what did you get

What’s the answer ?

I m getting W=v/l and option d.

And is A and C

Sure ?

Bcoz even on websites like Toppr and Doubtnut , they hve got W=v/l for exact same question