JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Laws of Motion

The electric force on the body is $qE=5\times 10^{-3}\times 200=1$ N, horizontal and pointing into the slope. It has a component $qE\cos 30^{\circ }$ up the incline and a component $qE\sin 30^{\circ }$ pressing the body against the surface, so the normal reaction is

and the friction is $\mu N=0.2\times 8.99=1.80$ N, acting up the slope against the sliding. Along the slope, with $m=1$ kg,

giving $a=2.23\mathrm{m}{\mathrm{s}}^{-2}$. The slope length for a drop of $1$ m is $s=\frac{1}{\sin 30^{\circ }}=2$ m, and starting from rest $s=\frac{1}{2}a{t}^2$ gives

The body reaches the bottom of the incline in $1.3$ s.