JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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If the earth is a point mass of

6 \times {10^{24}}\;kg

revolving around the sun at a distance of

1.5\; \times {10^8}\;km

and in time

T = 3.14 \times {10^7}s

, then the angular momentum of the earth around the sun is

View answer and explanation

Correct option: D

2.7 \times {10^{40}}\;kg\;{m^2}/s

Option D (

2.7 \times {10^{40}}\;kg\;{m^2}/s

) is correct because the angular momentum is calculated using the formula L = mvr = mr²ω, where m is the mass, r is the distance, and ω is the angular velocity.

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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)