JEE Main 2019 (Online) 10th April Morning Slot | Rotational Motion Question 108 | Physics

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

Two coaxial discs, having moments of inertia I₁ and I₁/2, are rotating with respective angular velocities $$\omega $$₁ and $$\omega $$₁/2 , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If E_f and E_i are the final and initial total energies, then (E_f - E_i) is:

$${{{I_1}\omega _1^2} \over {24}}$$

$${{{I_1}\omega _1^2} \over {12}}$$

$${3 \over 8}{I_1}\omega _1^2$$

$${{{I_1}\omega _1^2} \over {6}}$$

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

Moment of inertia of a body about a given axis is 1.5 kg m². Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular accleration of 20 rad/s² must be applied about the axis for a duration of :-

2.5 s

3 s

2 s

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

A thin smooth rod of length L and mass M is rotating freely with angular speed $$\omega $$₀ about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system , when the beads reach the opposite ends of the rod, will be :-

$${{M{\omega _0}} \over {M + 3m}}$$

$${{M{\omega _0}} \over {M + m}}$$

$${{M{\omega _0}} \over {M + 6m}}$$

$${{M{\omega _0}} \over {M + 2m}}$$

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $$\theta $$, where $$\theta $$ is the angle by which it has rotated, is given as k$$\theta $$². If its moment of inertia is I then the angular acceleration of the disc is :

$${k \over {4I}}\theta $$

$${k \over {I}}\theta $$

$${k \over {2I}}\theta $$

$${2k \over {I}}\theta $$

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