The radius of gyration of a disc of mass $100 g$ and radius $5 c m$ about an axis passing through its centre of gravity and perpendicular to the plane is (in cm)

0.5

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2.5

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3.54

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6.54

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Solution

The correct option is C $3.54$
Let radius of gyration be $k$
$∴ m k^{2} = m r^{2} k = \sqrt{\frac{R^{2}}{2}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = 3.54 c m$

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The radius of gyration of a disc of mass 100g and radius 5cm about an axis passing through its centre of gravity and perpendicular to the plane is (in cm)

The correct option is C 3.54Let radius of gyration be k∴mk2=mr2k=√R22=√252=5√2=3.54cm

The radius of gyration of a disc of mass $100 g$ and radius $5 c m$ about an axis passing through its centre of gravity and perpendicular to the plane is (in cm)

The correct option is C $3.54$
Let radius of gyration be $k$
$∴ m k^{2} = m r^{2} k = \sqrt{\frac{R^{2}}{2}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} = 3.54 c m$