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 R1 and R2 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
    $\frac{\text{R}_1}{\text{R}_2}$
  • C
    $\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$
  • D
    $\text{R}_1\text{R}_2.$

Answer

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

Explanation:

Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 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,

$\text{R}_2=\frac{\text{m}_2\text{v}_2}{\text{qB}}\Rightarrow\text{v}_2=\frac{\text{R}_2\text{qB}}{\text{m}_2}$

$\therefore\text{m}_1\Big(\frac{\text{R}_1\text{qB}}{\text{m}_1}\Big)^2=\text{m}_2\Big(\frac{\text{R}_2\text{qB}}{\text{m}_2}\Big)^2$

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

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