Four particles each of mass m are placed the corners of a square of side length l- The radius of gyration of the system about an axis perpendicular to the square and passing through its center is-ldfrac-l-sqrt-2-dfrac-l-2-sqrt-2- l

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The distance of mass from the center of square is r-l-x221A-2So the total moment of inertia of system I-4mr2-2ml2-xA0-Let the radius of gyration be kI-2ml2-4mk2-x21D2-k-l-x221A-2

Four particles each of mass $m$ are placed at the corners of a square of side length $l$ . The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:
$l$
$\frac{l}{\sqrt{2}}$
$\frac{l}{2}$
$(\sqrt{2}) l$

The distance of mass from the center of square is $r = \frac{l}{\sqrt{2}}$

So the total moment of inertia of system $I = 4 m r^{2} = 2 m l^{2}$

Let the radius of gyration be k

$I = 2 m l^{2} = 4 m k^{2} \Rightarrow k = \frac{l}{\sqrt{2}}$

Four particles each of mass m are placed at the corners of a square of side length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:ll√2l2(√2)l

The distance of mass from the center of square is r=l√2So the total moment of inertia of system I=4mr2=2ml2 Let the radius of gyration be kI=2ml2=4mk2⇒k=l√2

Four particles each of mass $m$ are placed at the corners of a square of side length $l$ . The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:
$l$
$\frac{l}{\sqrt{2}}$
$\frac{l}{2}$
$(\sqrt{2}) l$

The distance of mass from the center of square is $r = \frac{l}{\sqrt{2}}$

So the total moment of inertia of system $I = 4 m r^{2} = 2 m l^{2}$

Let the radius of gyration be k

$I = 2 m l^{2} = 4 m k^{2} \Rightarrow k = \frac{l}{\sqrt{2}}$