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DSSSB JE ME 2019 Official Paper Shift - 2 (Held on 06 Nov 2019)

Option 1 : 50 kg-m^{2}

__Concept:__

**Mass moment of inertia:**

I = M × K2

Where M = Mass of a body, K = Radius of gyration

The radius of gyration:

It is the radial distance from the point to the axis of rotation where the whole mass of the body is supposed to be concentrated so that the moment of inertia of the system is the same as the given situation.

Fig. A represents the moment of inertia for the body of mass M about the axis passing through it.

Fig. B represents the moment of inertia for point mass (having the same mass as the original body) placed at distance K from the axis of rotation.

I1 = I2

__Calculation:__

__Given:__

M = 2 kg, K = 5 m

I = M × K^{2} = 2 × 5^{2} = **50 kg-m2**

∴ The mass moment of inertia of the rigid body is 50 kg-m2