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

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

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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}}} $
$v_{p} : v_{d} : v_{\alpha}=1 : 1 : \sqrt{2}\;.$
answered Jan 29, 2014 by yamini.v
 

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