Question

A small block of mass $m$, charge $+q$ is kept at the top of a smooth inclined plane of angle ${30}^{\circ }$ placed in an elevator moving upward with an acceleration $a_0$.

Electric field $E$ exists 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).

• A. $t=\sqrt{\frac{2h}{g}}$
• B. $t=\sqrt{\frac{2h}{(g-a_0)+\frac{qE}{m}}}$
• C. $t=2\sqrt{\frac{2h}{(g+a_0)-\frac{\sqrt{3}qE}{m}}}$
• D. $t=\sqrt{\frac{2h}{(g+a_0{)}^2-\left(\frac{qE}{m}\right)h^2}}$

Answer 1

The correct option is C $t=2\sqrt{\frac{2h}{(g+a_0)-\frac{\sqrt{3}qE}{m}}}$

Net acceleration of the block down the incline is
$a=(g+a_0)\sin {30}^{\circ }-\frac{qE\cos {30}^{\circ }}{m}$
$a=\frac{(g+a_0)}{2}-\frac{\sqrt{3}qE}{2m}$

$\Rightarrow a=\frac{1}{2}(g+a_0-\frac{\sqrt{3}qE}{m})$

Now, length of the plane $l=\frac{h}{\sin 30}=\frac{1}{2}a{t}^2$

$\Rightarrow 2h=\frac{1}{4}[g+a_0-\sqrt{3}\left(\frac{qE}{m}\right)]t^2$

$\Rightarrow t=\sqrt{\frac{8h}{g+a_0-\sqrt{3}\left(\frac{qE}{m}\right)}}$

$\Rightarrow t=2\sqrt{\frac{2h}{g+a_0-\sqrt{3}\left(\frac{qE}{m}\right)}}$

Hence, the correct answer is (C).