Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$ . Then the ratio of $t_1$ to $t_2$ is:

$h_1 : h_2$

$\sqrt{h_1} : \sqrt{h_2}$

$h_1 : 2h_2$

$2h_1 : h_2$

Step-by-Step Solution

Equation of Motion: For an object falling freely from rest ( $u=0$ ) under gravity ( $a=g$ ), the distance covered $h$ in time $t$ is given by the second equation of motion: $h = ut + \frac{1}{2}gt^2$ $h = 0 + \frac{1}{2}gt^2 \implies h = \frac{1}{2}gt^2$
Relation between Time and Height: Rearranging the equation for time: $t = \sqrt{\frac{2h}{g}}$ Since $g$ is constant, $t \propto \sqrt{h}$ .
Calculate Ratio:

For height $h_1$ , time $t_1 = \sqrt{\frac{2h_1}{g}}$
For height $h_2$ , time $t_2 = \sqrt{\frac{2h_2}{g}}$
Ratio $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$ .

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