Question

Q.13 The radius of gyration of a disc of mass 100 gm and radius is 5 cm about an axis passing through center of gravity and perpendicular to the plane is A) 3.54 cm B) 1.54 cm​

Answer 1

Given :-

Mass of disc = M = 100g

Radius of disc = R = 5 cm

There is a relation between Radius of Gyration (k) and Moment of Inertia(I) i.e.

Moment Of Inertia of disc = I = MR²/2

I = Mk²

MR²/2 = Mk²

k² = (5)²/2

k = √12.5

k = 3.54 cm

Hence,

Radius of Gyration = k = 3.54 cm

So we can opt for (A).

Answer 2

Answer:

⇒ 3.54 is your answer

→ Solution:-

⇒                       I  = $\frac{MR^2}{2} (Disc)$

⇒                       I $= MK^2$

⇒ $\frac{MR^2}{2} = MK^2$

⇒ $\frac{R^2}{2} = K^2$

⇒ $K= \frac{R}{\sqrt{2} }$

⇒ $\frac{5cm}{\sqrt{2} } = 3.54cm$

⇒ $K = 3.54cm$

⇒Therefore , K is  A) 3.54cm.

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