JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Medium

1 of 1

Understanding +4 | -1

Assertion: The total distance moved by any point on the periphery of a wheel of radius

R

along the surface in one revolution is

2 \pi R.

Reason: In rolling motion of a wheel, every point on its periphery comes in contact with the surface once in one revolution.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the distance moved by a point on the wheel's periphery in one revolution is indeed

2 \pi R

, and this is because every point comes into contact with the surface once per revolution.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →

It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)