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A particle falls freely near the surface...

A particle falls freely near the surface of the earth. Consider a fixed point `O` (not vertically below the particle) on the ground.

A

Angular momentum of the particle about `O` is increasing

B

The moment of inertia of the particle about `O` is decreasing

C

The moment of inertia of the particle about `O` is decreasing

D

The angular velocity of the particle about `O` is increasing

Text Solution

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The correct Answer is:
B

The magnitude of angular momentum of the particle about `O=mvd`
Since speed `v` of the particle increases, its angular momentum about `O` increases.

Magnitude of inertia of the particle about `O=mr^(2)`
Hence `MI` of the particle about `O` decreases.
Angular velocity of the particle about `O=(vsintheta)/r`
Therfore `v` and `sintheta` increase and `r` decrease
Therefore, angular velocity of the particle about `O` increases.
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