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(D) When a particle of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = qV$

Vedclass · Accepted Answer

$\left[\frac{r_1}{r_2}\right]^2$

Vedclass · Answer

(D) When a particle of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.When this particle enters a uniform magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $r = \frac{mv}{qB}$.Substituting the expression for $v$,we get $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Since $q$,$V$,and $B$ are the same for both particles,$r \propto \sqrt{m}$,which means $r^2 \propto m$.Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left( \frac{r_1}{r_2} \right)^2$.

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

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