A particle has an angular momentum L and a linear momentum P about a reference point O- If r is the position vector of particle with respect to reference point- then identify correct option-

We know, L =r×P ∴L being the cross product vector, It will be nbsp; ⊥ to the plane containing both vectors nbsp; nbsp; P and r ⇒ Angle between vectors ...

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Standard XIII

Physics

Cross Product

A particle ha...

Question

A particle has an angular momentum $\to L$ and a linear momentum $\to P$ about a reference point $O$ . If $\to r$ is the position vector of particle with respect to reference point, then identify correct option.

\to L = \to P \times \to r

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| \to L \times \to P | = L P

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\to L \times \to P = 0

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None of these

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Solution

The correct option is B $| \to L \times \to P | = L P$
We know,
$\to L = \to r \times \to P$
$∴ \to L$ being the cross -product vector, It will be $⊥$ to the plane containing both vectors $\to P$ and $\to r$
$\Rightarrow$ Angle between vectors $\to L$ and $\to P$ will be $90^{\circ}$ .
$\Rightarrow | \to L \times \to P | = L P sin 90^{\circ} = L P$
However, $\to L . \to P = L P cos 90^{\circ} = 0$
$∴$ Option (b) is correct.

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Cross Product

Standard XIII Physics

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A particle has an angular momentum →L and a linear momentum →P about a reference point O. If →r is the position vector of particle with respect to reference point, then identify correct option.

A particle has an angular momentum $\to L$ and a linear momentum $\to P$ about a reference point $O$ . If $\to r$ is the position vector of particle with respect to reference point, then identify correct option.