Tesla is the unit of

$A$ positive,singly ionized atom of mass number $A_M$ is accelerated from rest by a voltage $V = 192 \text{ V}$. Thereafter,it enters a rectangular region of width $w$ with a magnetic field $B_0 = 0.1 \hat{k} \text{ T}$,as shown in the figure. The ion finally hits a detector at a distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $= (5/3) \times 10^{-27} \text{ kg}$,charge of the electron $= 1.6 \times 10^{-19} \text{ C}$.]
Which of the following option$(s)$ is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ ion is $4 \text{ cm}$.
$(B)$ The value of $x$ for an ion with $A_M = 144$ is $48 \text{ cm}$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$,the minimum height $(x_1 - x_0)$ of the detector is $52 \text{ cm}$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $28 \text{ cm}$.

An electron having mass $9.1 \times 10^{-31} \ kg$,charge $1.6 \times 10^{-19} \ C$ and moving with velocity of $10^6 \ ms^{-1}$ enters a region where a magnetic field exists. If it describes a circle of radius $0.2 \ m$,then the intensity of the magnetic field must be . . . . . . $\times 10^{-5} \ T$.

Two particles $X$ and $Y$ having equal charges,are accelerated through the same potential difference. They enter a region of a uniform magnetic field and describe a circular path of radii $r_1$ and $r_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

If a charged particle enters perpendicularly in a uniform magnetic field,then:

An electron with energy $0.1 \, keV$ moves at a right angle to the Earth's magnetic field of $1 \times 10^{-4} \, Wb/m^2$. The frequency of revolution of the electron will be. (Take mass of electron $= 9.0 \times 10^{-31} \, kg$)