A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be
mg – T = ma. (1)

T * R = l α. (2)
a = α R. (3)
With the help of equations (1), (2) and (3), we get
Similar Questions for you
mg = 10 × 10
= 100 of
NA = reaction force offered by the wall
reaction force offered by the floor.
By NLM 1
Translational fB = NA
NB = mg = 100N
Ac2 + Bc2 = AB2
Ac2 = 34 – 9
Ac = 5m
Rotational equilibrium
mg
NA = 50 cot
Now cot =
NA =
Therefore, NA = - (i)
-
a = 6t2 – 2t, of t 20, w = 10 rad/sec
, q = 4 rad
=
A/c to question
f + 76 = 2f
f = 76
last frequency
= f + 76 = 2 × 76
= 152 H2
K.ER (Rotational) = - (1)
- (2)
At maximum height velocity (v) = 0
So, momentum = mv = m × 0 = 0
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Physics NCERT Exemplar Solutions Class 12th Chapter Seven 2025
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