The correct option is A 25 rad s−2
Given, m=3 kg, r=40 cm, F=30 N
Moment of inertia of hollow cylinder about its axis
=mr2=3×(0.4)2=0.48 kg m2
The torque is given by,
τ=Ia
Where, I= moment of inertia and a = angular accleration
In the given case
τ=rF, as the force is acting perpendicularly to the radial vector.
∴a=τI=Frmr2=Fmr=303×40×10−2a= 25 rad s−2