If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be
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(c)
$\Delta K E=q V$
$\frac{\Delta KE _\alpha}{\Delta KE _p}=\frac{q_\alpha V}{q_p V}=\frac{q_\alpha}{q_p}=2$
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