Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Two balls are dropped from heights h and 2h respectively from the earth surface. The ratio of time of these balls to reach the earth is:

1 : \sqrt{2}

\sqrt{2} : 1

2 : 1

1 : 4

Principle: For a body in free fall starting from rest ( $u=0$ ), the distance fallen ( $h$ ) in time ( $t$ ) is given by the kinematic equation: $h = ut + \frac{1}{2}gt^2 \Rightarrow h = \frac{1}{2}gt^2$
Relation: This implies that time is proportional to the square root of the height: $t = \sqrt{\frac{2h}{g}} \Rightarrow t \propto \sqrt{h}$ .
Calculation: For the first ball ( $h_1 = h$ ): $t_1 \propto \sqrt{h}$ For the second ball ( $h_2 = 2h$ ): $t_2 \propto \sqrt{2h}$
Ratio: $\frac{t_1}{t_2} = \frac{\sqrt{h}}{\sqrt{2h}} = \frac{1}{\sqrt{2}}$ The ratio is $1 : \sqrt{2}$ .

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