If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane as shown in fig. Its angular momentum with respect to origin at any time t will be

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Solution

The correct option is B
We know that, Angular momentum,
$\to L$ = $\to r$ $\times$ $\to p$ in terms of component becomes,
$\to L$ = $⎡ ⎢ ⎢ ⎢ ⎢ ⎢ \begin{matrix} ^i & ^j & ^k x & y & z p_{x} & p_{y} & p_{z} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥$

As motion is in x-y plane(z = 0 and $P_{z}$ = 0), so $\to L$ = $\to k$ (x $p_{y} + y p_{x}$ )
Here x = vt, y = b, $p_{x}$ = mv and $p_{y}$ = 0
$∴$ $\to L$ = $\to k$ [vt $\times$ 0 - bmv] = -mvb $^k$

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