Angular momentum of a single particl moving with constant speed along circular path:

Correct Option - 3


Detailed Solution:

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}\Rightarrow \overrightarrow{L}=mvr\left(\widehat{k}\right)$

Direction & magnitude both remain same

 for particle moving with constant speed.

<p><span contenteditable="false"> <math> <mrow> <mover accent="true"> <mi>L</mi> <mo>→</mo> </mover> <mo>=</mo> <mover accent="true"> <mi>r</mi> <mo>→</mo> </mover> <mo>×</mo> <mover accent="true"> <mi>p</mi> <mo>→</mo> </mover> <mo>⇒</mo> <mover accent="true"> <mi>L</mi> <mo>→</mo> </mover> <mo>=</mo> <mi>m</mi> <mi>v</mi> <mi>r</mi> <mrow> <mo> (</mo> <mrow> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754454014phpfKkD57.jpeg" alt="" width="90" height="70" loading="lazy"></picture></div></div><p>Direction &amp; magnitude both remain same</p><p>&nbsp;for particle moving with constant speed.</p>

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