A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top

The correct option is 2 s.

When a body slides down a smooth inclined plane, it moves with a constant acceleration ($a$). The distance ($s$) covered by the body starting from rest (initial velocity $u=0$) in time ($t$) is given by the second equation of motion:
$s=ut+\frac{1}{2}a{t}^2$
Since the body starts from rest, the equation simplifies to:
$s=\frac{1}{2}a{t}^2$
This shows that the distance covered is directly proportional to the square of the time taken ($s\propto t^2$).

Let $d$ be the total distance of the inclined plane and $t_1$ be the total time taken to cover this distance.
Given:
Total time, $t_1=4\text{s}$
For the total distance:
$d=\frac{1}{2}a{t}_1^2$ (Equation 1)

Let $t_2$ be the time taken to cover one-fourth of the distance ($\frac{d}{4}$).
For this partial distance:
$\frac{d}{4}=\frac{1}{2}a{t}_2^2$ (Equation 2)

Dividing Equation 2 by Equation 1:
$\frac{d/4}{d}=\frac{\frac{1}{2}a{t}_2^2}{\frac{1}{2}a{t}_1^2}$
Simplifying the relation:
$\frac{1}{4}=\frac{t_2^2}{t_1^2}$
Taking the square root on both sides:
$\frac{1}{2}=\frac{t_2}{t_1}$
Solving for $t_2$:
$t_2=\frac{t_1}{2}$

Substitute the value of $t_1=4\text{s}$:
$t_2=\frac{4}{2}=2\text{s}$
Therefore, it takes 2 seconds to cover one-fourth of the distance.