Test: Kinematics & Dynamics of Rotational Motion about a fixed Axis (18 July) - JEE MCQ

If an object is rolling without slipping, then its kinetic energy can be expressed as the sum of the translational kinetic energy of its center of mass plus the rotational kinetic energy about the center of mass.

Newton's First Law for Rotation: an object at rest tends to remain at rest, and an object that is spinning tends to spin with a constant angular velocity, unless it is acted on by a nonzero net torque or there is a change in the way the object's mass is distributed.

In the fixed axis rotation we see that every point on the body has two components of velocity, one in the radial direction and one in the tangential direction. The resultant of these velocities is not the same for any two points lying in the plane of the body. 
Any two points on the radial line have the radial acceleration directed towards the center of equal magnitude and the tangential acceleration of equal magnitude as well. Thus option B is correct.
All the particles lying on the curved surface of a cylinder whose axis coincides with the axis of rotation have the same speed but different velocities.

The answer is option A
So, as we know,
After differentiating angular velocity with respect to t,
Linear velocity=2t+2.3=ω
Now,
Velocity=r x ω(where r is 0.8m)
=0.8x2.3
=1.84m/s

Angular displacement is measured in units of radians. 2 pi radians equals 360 degrees. The angular displacement is not a length (i.e. not measured in meters or feet), so an angular displacement is different than a linear displacement.

r=0.5m, t=15s
n₁=5rpm=5/60 rps
n₂=25rpm=25/60rps
… ω₁=2π(5) rad/s
… ω₂=2π(25) rad/s
As,
a=r∝          [∝=ω₂- ω₁/t]
so, a=0.5[2π(n₂-n₁)/t]
a=0.5x6.28x20/60x15
a=6.28x2/60x3
a=6.28/90
a=0.697 m/s²
a≈0.7 m/s²