JEE Main 2019 (Online) 10th April Morning Slot | Rotational Motion Question 140 | Physics

A particle of mass m is moving along a trajectory given by
x = x₀ + a cos$$\omega $$₁t
y = y₀ + b sin$$\omega $$₂t
The torque, acting on the particle about the origin, at t = 0 is :

Zero

+my₀a $$\omega _1^2$$$$\widehat k$$

$$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$$

m (–x₀b + y₀a) $$\omega _1^2$$$$\widehat k$$

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

The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane. : (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radius R/4 . If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is :

20 : 15 : 14

4 : 3 : 2

2 : 3 : 4

10 : 15 : 7

PREVIOUS NEXT

Questions Asked from Rotational Motion (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions