Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 2
PhysicsNumerical+2Question 10 of 57

JEE Advanced 2021 · Paper 2, Physics question 10

Passage

Question Stem for Question Nos. 9 and 10

A pendulum consists of a bob of mass m=0.1m = 0.1 kg and a massless inextensible string of length L=1.0L = 1.0 m. It is suspended from a fixed point at height H=0.9H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P=0.2P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is JJ kg-m2^2/s. The kinetic energy of the pendulum just after the lift-off is KK Joules.

The value of KK is ____.

Answer and worked solution

Official answer: 0.14

When the string becomes taut. It makes angle θ\theta with the vertical, where cos⁡θ=0.91.0\cos\theta=\frac{0.9}{1.0}. The impulsive tension removes the velocity component along the string. Only the component perpendicular to it survives:
v⊥=vcos⁡θ=2×0.9=1.8 m/s.v_\perp=v\cos\theta=2\times0.9=1.8\text{ m/s}.

Check. Angular momentum about the suspension point is conserved during the jerk: mv⊥L=0.1×1.8×1=0.18=Jmv_\perp L=0.1\times1.8\times1=0.18=J.

Kinetic energy just after lift-off.
K=12(0.1)(1.8)2=0.162 J.K=\tfrac12(0.1)(1.8)^2=0.162\text{ J}.

Topics:Impulse and momentumAngular momentum and its conservationConservation of mechanical energy
← Previous questionNext question →