Ans : (d)
If V is the accelerating potential and v is the velocity of electron, then
$eV = \large\frac{1}{2} mv^2$
$v \sim \sqrt V$
$ \Rightarrow \large\frac{v_1}{v_2} = \sqrt { \bigg( \large\frac{V_1}{V_2} \bigg) }$
Here, $V_1 = V, V_2 = 2V; v_1 = v$
So, $ \large\frac{v}{v_2} = \sqrt { \large\frac{V}{2V} }= \large\frac{1}{ \sqrt 2}$
$ \Rightarrow v_2 = \sqrt 2v = 1.4v$