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The angular momentum of a particle about $o$ is varying as $L = 4t+8$ (SI units) . When it moves along straight line $y= x-4$(x,y in maters) . The magnitude of force acting on the particle would be

\[\begin {array} {1 1} (a)\;1N & \quad (b)\;2N \\ (c)\;\sqrt 2 N & \quad  (d)\;\sqrt 3 N \end {array}\]
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$OB= +4 (y=0 => x=4)$
$slope \;=1$
$\angle OBC=45$
$AB= 4 \sqrt 2$
$AC= 2 \sqrt 2=CB$
In $\Delta OCB, \tan 45=\large\frac{OC}{BC}$$=1$
$OC=2 \sqrt 2$
$\large\frac{d \bar {l}}{dt}$$= \overrightarrow{ r} \times \overrightarrow {F}$
$4 =rF$
$4= 2 \sqrt 2.F$
$F= \sqrt 2 N$
answered Nov 28, 2013 by meena.p
edited Jun 22, 2014 by lmohan717

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