A particle of mass m and charge q enters a region of uniform magnetic field vec B with speed v, its velocity making an angle of 60^° with vec B. A…

Perpendicular to $\vec B$ the motion is circular with speed $v\sin60^\circ=\tfrac{\sqrt3}{2}v$, so $r=\dfrac{m(\sqrt3 v/2)}{qB}=\dfrac{\sqrt3mv}{2qB}$; the electric field does not alter this, so (A) is correct. The period $T=2\pi m/qB$ is independent of speed, so (B) is correct. Along $\vec B$ the initial speed is $v\cos60^\circ=v/2$ and the acceleration is $qE/m=qv B/(2\pi m)$. In time $T$: $z=\tfrac v2T+\tfrac12\dfrac{qE}{m}T^2=\dfrac{\pi mv}{qB}+\dfrac{\pi mv}{qB}=\dfrac{2\pi mv}{qB}$, so (C) is correct. The work done by the electric field is $qEz=q\cdot\dfrac{vB}{2\pi}\cdot\dfrac{2\pi mv}{qB}=mv^2$, which is twice the initial kinetic energy $\tfrac12mv^2$. The final kinetic energy is therefore $3\times$ the initial value, not $2\times$, so (D) is incorrect. Hence (A), (B), (C).