JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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1 of 1

Application +4 | -1

Let

\mathrm{M}

and

\mathrm{L}

be the mass and length of thin uniform rod respectively. In

1^{\text {st }}

case, axis of rotation is passing through centre and perpendicular to its length. In

2^{\text {nd }}

case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is

View answer and explanation

Correct option: A

1: 2

Option A (1:2) is correct because the ratio of the radius of gyration is calculated based on the moment of inertia for each case.

More Questions from SYSTEM OF PARTICLES

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)