A particle of mass 100,g is projected at time | vedclass

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(B) The angular momentum $L$ of a particle about the origin is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.Alternatively,the tor

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$800$

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(B) The angular momentum $L$ of a particle about the origin is given by $L = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.Alternatively,the torque about the origin is $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes (m\vec{g})$.Since $\vec{g}$ acts downwards ($-y$ direction),$\vec{	au} = (x\hat{i} + y\hat{j}) 	imes (-mg\hat{j}) = -xmg\hat{k}$.The magnitude of torque is $	au = xmg = (v_x t)mg$.Integrating torque to find angular momentum: $L = \int_0^t 	au dt = \int_0^t (v_x t)mg dt = mg v_x \frac{t^2}{2}$.Given: $m = 100\,g = 0.1\,kg$,$v = 20\,ms^{-1}$,$	heta = 45^{\circ}$,$t = 2\,s$,$g = 10\,ms^{-2}$.$v_x = v \cos 45^{\circ} = 20 	imes \frac{1}{\sqrt{2}} = 10\sqrt{2}\,ms^{-1}$.$L = (0.1)(10)(10\sqrt{2}) \frac{2^2}{2} = 10 	imes 10\sqrt{2} 	imes 2 = 200\sqrt{2} = \sqrt{40000 	imes 2} = \sqrt{80000}$.Wait,re-evaluating: $L = (0.1)(10)(10\sqrt{2}) 	imes 2 = 20\sqrt{2} = \sqrt{400 	imes 2} = \sqrt{800}$.Thus,$K = 800$.

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