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MHT CET 2026 (17 Apr, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2026 (17 Apr, Shift 2)

Q11MHT CET 2026 · 17 APR · SHIFT 2Medium

Angular Momentum & its Conservation

A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is 'L'. If the length of the string is now halved, keeping angular velocity same then the angular momentum will be

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Answer: (a)

$L=I\omega=mr^2\omega$
$r\to\frac r2$ with same $\omega$ : $L'=m\frac{r^2}{4}\omega$
$L'=\frac L4$
So the answer is (a) $\frac L4$ .

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