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(B) The radius $R$ of a circular path for a charged particle moving in a magnetic field $B$ is given by $R = \frac{mv}{qB}$.Since kinetic energy $K =

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$5:4$

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(B) The radius $R$ of a circular path for a charged particle moving in a magnetic field $B$ is given by $R = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Substituting this into the radius formula,we get $R = \frac{\sqrt{2mK}}{qB}$.Rearranging for charge $q$,we get $q = \frac{\sqrt{2mK}}{RB}$.Given that $K$ and $B$ are the same for both particles,the ratio of charges is $\frac{q_1}{q_2} = \sqrt{\frac{m_1}{m_2}} 	imes \frac{R_2}{R_1}$.Given $\frac{R_1}{R_2} = \frac{6}{5}$ and $\frac{m_1}{m_2} = \frac{9}{4}$,we have $\frac{q_1}{q_2} = \sqrt{\frac{9}{4}} 	imes \frac{5}{6} = \frac{3}{2} 	imes \frac{5}{6} = \frac{15}{12} = \frac{5}{4}$.

Two charged particles,having the same kinetic energy,are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of the radii of their circular paths is $6:5$ and their respective masses ratio is $9:4$,then the ratio of their charges will be.

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