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(B) The centripetal force is provided by the magnetic force.$i.e., \frac{mv^2}{R} = qvB ......... (1)$where $m = 	ext{mass of the ion}$,$v = 	ext{ve

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$\frac{1}{R^2}$

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(B) The centripetal force is provided by the magnetic force.$i.e., \frac{mv^2}{R} = qvB ......... (1)$where $m = 	ext{mass of the ion}$,$v = 	ext{velocity}$,$q = 	ext{charge of ion}$,$B = 	ext{magnetic field flux density}$.From $(1)$,the velocity of the ion is $v = \frac{qBR}{m}$.The kinetic energy gained by the ion when accelerated through a potential $V$ is given by $E = qV$.Also,$E = \frac{1}{2}mv^2$.Equating the two expressions for energy:$qV = \frac{1}{2}m \left( \frac{qBR}{m} ight)^2$$qV = \frac{1}{2}m \frac{q^2 B^2 R^2}{m^2}$$qV = \frac{q^2 B^2 R^2}{2m}$Rearranging for the ratio $\frac{q}{m}$:$\frac{q}{m} = \frac{2V}{B^2 R^2}$Since $V$ and $B$ are constant,we have $\frac{q}{m} \propto \frac{1}{R^2}$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

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