
$\mathrm{qV}=\frac{1}{2} \mathrm{mv}_{0}^{2}$
$v_{0}=\sqrt{\frac{2 q V}{m}}$
The radius of circular path in magnetic Field is
$r=\frac{m v_{0}}{q B}$
or $\frac{\mathrm{d}}{2}=\frac{\mathrm{mv}_{0}}{\mathrm{qB}}$
or $\frac{d^{2}}{4}=\frac{m^{2} v_{0}^{2}}{q^{2} B^{2}}=\frac{m^{2}}{q^{2} B^{2}} \frac{2 q V}{m}$
or $\frac{d^{2}}{4}=\frac{2 m V}{q B^{2}}$
$\therefore \quad \mathrm{m}=\frac{\mathrm{qB}^{2} \mathrm{d}^{2}}{8 \mathrm{V}}$

$(A)$ $\rho_A>\rho_B$ and $m_A=m_B$
$(B)$ $\rho_A<\rho_B$ and $m_A=m_B$
$(C)$ $\rho_A>\rho_B$ and $m_A > m_B$
$(D)$ $\rho_A<\rho_B$ and $m_A < m_B$
