A small block of mass m, charge +q is kept at the top of a smooth inclined plane of angle 30° placed in an elevator moving upward with an acceleration a_0.Electric field E exits between the vertical side walls of the elevator. The time taken by the block to come to the lowest point of inclined plane is ( assuming the surface to be smooth).

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$t = 2 \sqrt{\frac{2 h}{(g + a_{0}) - \sqrt{3} \frac{qE}{m}}}$

$t = \sqrt{\frac{2 h}{{(g + a_{0})}^{2} - (\frac{qE}{m}) h^{2}}}$

$t = \sqrt{\frac{2 h}{(g - a_{0}) + \frac{qE}{m}}}$

$t = \sqrt{\frac{2 h}{g}}$

answer is C.

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Detailed Solution

Net acceleration of block down the incline is
$a = \frac{(g + a_{0})}{2} - \frac{qE \sqrt{3}}{2 m} \Rightarrow a = \frac{1}{2} (g + a_{0} - \frac{qE \sqrt{3}}{m})$

$Now, \frac{h}{\sin 30} = \frac{1}{2} {at}^{2} \Rightarrow 2 h = \frac{1}{4} [g + a_{0} - \sqrt{3} (\frac{qE}{m})] t^{2} \Rightarrow t = \sqrt{\frac{8 h}{g + a_{0} - (\frac{qE}{m}) \sqrt{3}}} \Rightarrow t = 2 \sqrt{\frac{2 h}{(g + a_{0}) - (\frac{qE}{m}) \sqrt{3}}}$