If charge is moving perpendicular to uniform magnetic field then its time period of revolution is:

independent of its radius of circular path

The correct answer is Option (1) → independent of its radius of circular path

The time period (T) of revolution is -

$T=\frac{2πr}{v}=\frac{2π.\frac{mv}{qB}}{v}$

$⇒T=\frac{2πm}{qB}$

Hence, it is independent of it's Radius.