JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A ball is spun with angular acceleration

\alpha = 6{t^2} - 2t

, where

t

is in second and

\alpha

is in

rad{s^{ - 2}}

. At

t = 0

, the ball has angular velocity of

10\,rad{s^{ - 1}}

and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

View answer and explanation

Correct option: B

\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4

Option B (

\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4

) is correct because it is obtained by integrating the angular acceleration twice with respect to time, using the given initial conditions for angular velocity and angular position.

More Questions from WORK-ENERGY-POWER

Practice All Chapter MCQs →

A ball throwing cannon, lifts ball and throws with

72 \mathrm{~km} / \mathrm{hr}

Speed. If balls are raised upto

10 \mathrm{~m}

. then find Input power of cannon if it lifts 8 balls per minute and its efficiency is

80 \%

. Mass of each ball is

1 \mathrm{~kg}

. Assertion:When a particle moves in a circle with a uniform speed its acceleration is constant but the velocity changes.
Reason: Angular displacement is not an axial vector. A body of mass

m

thrown vertically upwards attains a maximum height

h

. As what height will its kinetic energy be

75 \%

of its initial value? Assertion: Centripetal force does no work.

Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.

Statement II: Work done against restoring force is stored as potential energy. A bullet of mass

m

moving at a speed

v

hits a ball of mass

M

kept at rest. A small part having mass

m^{\prime}

breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed

v_1

in the direction of the bullet. Find the velocity of the bullet after the collision.