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(B) The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicul

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$60$

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(B) The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicular distance from the origin to the line of motion.The equation of the line is $2x - y + 4 = 0$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = 2$,$B = -1$,and $C = 4$.$d = \frac{|4|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{4 + 1}} = \frac{4}{\sqrt{5}} \ m$.Given mass $m = 5 \ kg$ and velocity $v = 3\sqrt{5} \ m/s$.$L = mvd = 5 	imes (3\sqrt{5}) 	imes \left(\frac{4}{\sqrt{5}}ight)$.$L = 5 	imes 3 	imes 4 = 60 \ kg \ m^2 \ s^{-1}$.

$A$ particle of mass $5 \ kg$ moves along the straight line $y = 2x + 4$ with a velocity of $3\sqrt{5} \ m/s$. Find its angular momentum about the origin in $kg \ m^2 \ s^{-1}$.

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