# A proton , a deutron and alpha particle are accelerated through potentials of $\;V , 2V ,4V\;$respectively . Their velocity will bear a ratio .

$(a)\;1 :1 :1\qquad(b)\;1 :\sqrt{2} : 1\qquad(c)\;\sqrt{2} : 1: 1\qquad(d)\;1 : 1 : \sqrt {2}$

Answer : (d) $\;1 : 1 : \sqrt{2}$
Explanation :
$\large\frac{1}{2}m_{p}v_{p}^2=V e _{p}\quad\;v_{p}=\sqrt{\large\frac{2V e_{p}}{m_{p}}}$
$\large\frac{1}{2}(2m_{p})v_{d}^2=V e _{p}\quad\;v_{d}=\sqrt{\large\frac{2V e_{p}}{m_{p}}}$
$\large\frac{1}{2}(4m_{p})v_{\alpha}^2=4V \times\;2e _{p}\quad\;v_{d}=\sqrt{2}\sqrt{\large\frac{2V e_{p}}{m_{p}}}$
$v_{p} : v_{d} : v_{\alpha}=1 : 1 : \sqrt{2}\;.$