Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to the magnetic field?

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

If a charged particle moves in a gravity-free space with uniform velocity,then which of the following is not possible? ($\overrightarrow{E} =$ electric field,$\overrightarrow{B} =$ magnetic field)

$A$ particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{i}$ with an initial velocity $\vec{v} = v_0 \hat{j}$. The speed of the particle will become $2v_0$ after a time:

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is