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

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2 N

$\sqrt{3} N$

1 N

$\sqrt{2} N$

answer is C.

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Detailed Solution

$τ = \frac{d L}{d t} = 4$

y=x-4

If x=0 then y=-4

If x=4 then y=0

slope =1

From figure $r_{⊥} = 2 \sqrt{2}$

Hence $τ = r_{⊥} F$

$4 = 2 \sqrt{2} . F$ $∴ F = \sqrt{2} N$

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