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(D) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$.Since $v = r\omega$ and $\omega = \frac{2\pi}{T}$,we have $L = m(r

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$2.7 	imes 10^{40} \ kg \ m^2/s$

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(D) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$.Since $v = r\omega$ and $\omega = \frac{2\pi}{T}$,we have $L = m(r\omega)r = m\omega r^2 = m \left( \frac{2\pi}{T} ight) r^2$.Given: $m = 6 	imes 10^{24} \ kg$,$r = 1.5 	imes 10^8 \ km = 1.5 	imes 10^{11} \ m$,$T = 3.14 	imes 10^7 \ s$.Substituting the values:$L = \frac{6 	imes 10^{24} 	imes 2 	imes 3.14 	imes (1.5 	imes 10^{11})^2}{3.14 	imes 10^7}$$L = \frac{6 	imes 10^{24} 	imes 2 	imes 2.25 	imes 10^{22}}{10^7}$$L = 27 	imes 10^{39} = 2.7 	imes 10^{40} \ kg \ m^2/s$.

If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

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