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(C) The kinetic energy gained by the electron is given by $K.E. = \frac{1}{2}mv^2 = eV$.From this,the velocity $v$ is $v = \sqrt{\frac{2eV}{m}}$.The m

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$4: 1$

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(C) The kinetic energy gained by the electron is given by $K.E. = \frac{1}{2}mv^2 = eV$.From this,the velocity $v$ is $v = \sqrt{\frac{2eV}{m}}$.The magnetic force $F$ on an electron moving with velocity $v$ in a magnetic field $B$ is $F = evB \sin(	heta)$. Assuming the velocity is perpendicular to the magnetic field,$F = evB$.Substituting the value of $v$: $F = eB \sqrt{\frac{2eV}{m}} = B \sqrt{\frac{2e^3V}{m}}$.This shows that $F \propto \sqrt{V}$.Therefore,$\frac{F_2}{F_1} = \sqrt{\frac{V_2}{V_1}}$.Given $\frac{F_2}{F_1} = \frac{2F}{F} = 2$.Squaring both sides,we get $\frac{V_2}{V_1} = (2)^2 = 4$.Thus,the ratio $\frac{V_2}{V_1} = 4:1$.

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