$A$ particle is projected with a velocity of $10 \ m/s$ along the $y-$axis from the point $(2, 3)$. $A$ uniform magnetic field of $(3\hat{i} + 4\hat{j}) \ T$ exists in the space. What is its speed when the particle passes through the $y-$axis for the third time? (Neglect gravity)

Consider the following statements regarding a charged particle in a magnetic field. Which of the statements are true?

$A$ 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:

$A$ charge moving with velocity $v$ in $X$-direction is subjected to a field of magnetic induction in the negative $X$-direction. As a result,the charge will

$A$ nucleus at rest undergoes $\alpha -$ decay according to the equation ${}_{92}^{226}X \to Y + \alpha$. At $t=0$,the emitted $\alpha -$ particle enters a region of space where a uniform magnetic field $\vec{B} = B_0 \hat{i}$ and an electric field $\vec{E} = E_0 \hat{i}$ exist. The $\alpha -$ particle enters the region with velocity $\vec{v} = v_0 \hat{j}$ at $x = 0$. After a time $t = \left( \sqrt{3} \times 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} \right)$,the particle is found to have twice its initial speed. The initial speed $v_0$ of the $\alpha -$ particle is:

$A$ particle having a charge of $10.0\,\mu C$ and mass $1\,\mu g$ moves in a circle of radius $10\,cm$ under the influence of a magnetic field of induction $0.1\,T$. When the particle is at a point $P$, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is......$V/m$