The radius of gyration of a disc about its axis passing through its centre and perpendicular to its plane is

R / \sqrt 2

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R / 2

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\sqrt 2 R

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2 R

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Solution

The correct option is A $R / \sqrt 2$
Moment of Inertia of disc about center of mass is $I = \frac{m R^{2}}{2}$
Let a line parallel to axis at a distance $d$ .
Radius of Gyration $= R$
Then, $\frac{m R^{2}}{2} = m R^{2}$
$\Rightarrow R^{2} = \frac{R^{2}}{2}$
$\Rightarrow R = \frac{R}{\sqrt{2}}$

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The radius of gyration of a disc about its axis passing through its centre and perpendicular to its plane is

The correct option is A R/√2Moment of Inertia of disc about center of mass is I=mR22Let a line parallel to axis at a distance d.Radius of Gyration=RThen, mR22=mR2⇒R2=R22⇒R=R√2

The correct option is A $R / \sqrt 2$
Moment of Inertia of disc about center of mass is $I = \frac{m R^{2}}{2}$
Let a line parallel to axis at a distance $d$ .
Radius of Gyration $= R$
Then, $\frac{m R^{2}}{2} = m R^{2}$
$\Rightarrow R^{2} = \frac{R^{2}}{2}$
$\Rightarrow R = \frac{R}{\sqrt{2}}$