JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

If the earth is a point mass of

6 \times {10^{24}}

kg revolving around the sun at distance of

1.5 \times {10^8}

km and in time T =

3.14 \times {10^7}

seconds, then the angular momentum of the earth around the sun is [in

kg - {m^2}

/ s]

View answer and explanation

Correct option: D

2.7 \times {10^{40}}

Option D (

2.7 \times {10^{40}}

) is correct because it is the angular momentum calculated using the formula L = mvr = mωr².

More Questions from SYSTEM OF PARTICLES

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\mathrm{AB}

\mathrm{L}

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

x

\mathrm{A}

\mathrm{CM}

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

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

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

t = 1\,sec

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