Two particles X and Y having equal charge, after being accelerate… - Vidyadip

MCQ

Two particles $X$ and $Y$ having equal charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is:

A
$\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^{\frac{1}{2}}$
B
$\frac{\text{R}_1}{\text{R}_2}$
✓
$\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$
D
$\text{R}_1\text{R}_2.$

✓

Answer

Correct option: C.

$\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$

Particles $X$ and $Y$ of respective masses $m_1$ and $m_2$ are carrying charge $q$ describing circular paths with respective radii $R_1$ and $R_2$ such that,
$\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}$
$\text{R}_1=\frac{\text{m}_2\text{v}_2}{\text{qB}}$
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
$\therefore\frac{1}{2}\text{m}_1\text{v}_1^2=\frac{1}{2}\text{m}_2\text{v}_2^2$
$\because\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}\Rightarrow\text{v}_1=\frac{\text{R}_1\text{qB}}{\text{m}_1}$
And,
$\text{R}_2=\frac{\text{m}_2\text{v}_2}{\text{qB}}\Rightarrow\text{v}_2=\frac{\text{R}_2\text{qB}}{\text{m}_2}$
$\therefore\text{m}_1\Big(\frac{\text{R}_1\text{qB}}{\text{m}_1}\Big)^2=\text{m}_2\Big(\frac{\text{R}_2\text{qB}}{\text{m}_2}\Big)^2$
$\Rightarrow\frac{\text{m}_1}{\text{m}_2}=\frac{\text{R}_1^2}{\text{R}_2^2}=\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Magnetic Field due to a Current→Magnetic Field due to a Current — all question types→Physics STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Energy of electrons can be increased by allowing them(a) To fall through electric potential(b) To move in high magnetic field(c) To fall from great heights(d) To pass through lead blocks

Zener breakdown takes place if(a) Doped impurity is low(b) Doped impurity is high(c) Less impurity in N-part(d) Less impurity in P-type

Relation between current (i) and drift velocity $\left(v_d\right)$ is :

A permanent magnet(a) Attracts all substances(b) Attracts only magnetic substances(c) Attracts magnetic substances and repels all non-magnetic substances(d) Attracts non-magnetic substances and repels magnetic substances

The velocity of X-rays having wavelength 1 Å will be :

In a triode valve the amplification factor is $ 20$ and mutual conductance is $10^{–3}$ mho. The plate resistance is

If the mass of neutron $=1.7\times 10−27\ kg .$ Then the de broglie wavelength of neutron of energy $3\text{eV}$ is:

A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and another point Q lies to the west in the same direction. The magnetic field at P is:

In an $NPN$ transistor circuit, the collector current is $10 mA$. If $90\%$ of the electrons emitted reach the collector, the emitter current $(i_E)$ and base current $(i_B)$ are given by

The magnetic flux linked with a vector area $\overrightarrow{\mathrm{A}}$ in a uniform magnetic field $\overrightarrow{\mathrm{B}}$ is