Uniform magnetic fields of different strengths $\left(B_1\right.$ and $\left.B_2\right)$, both normal to the plane of the paper exist as shown in the figure. A charged particle of mass $m$ and charge $q$, at the interface at an instant, moves into the region 2 with velocity $v$ and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?
(Consider the velocity of the particle to be normal to the magnetic field and $\mathrm{B}_2>\mathrm{B}_1$ )
Given below are two statements: one is labelled as $\mathbf{A s s e r t i o n} \mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A : If Oxygen ion $\left(\mathrm{O}^{-2}\right)$ and Hydrogen ion $\left(\mathrm{H}^{+}\right)$enter normal to the magnetic field with equal momentum, then the path of $\mathrm{O}^{-2}$ ion has a smaller curvature than that of $\mathrm{H}^{+}$.
Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statements, choose the correct answer from the options given below
In a moving coil galvanometer, two moving coils $\mathrm{M}_1$ and $\mathrm{M}_2$ have the following particulars :
$$ \begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \\ & \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T} \end{aligned} $$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?