Q. A particle is dropped vertically from rest a height. The time taken by it to fall through successive distances of $1\ m$ each will then be
A) All equal, being equal to $\sqrt{2/g}$ second
B) In the ratio of the square roots of the integers $1,\ 2,\ 3$…….
C) In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1},\ \left( \sqrt{2} - \sqrt{1} \right),\left( \sqrt{3} - \sqrt{2} \right),\left( \sqrt{4} - \sqrt{3} \right)$……
D) In the ratio of the reciprocal of the square roots of the integers i.e.,. $\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{4}}$
Explanation
(c) $h = ut + \frac{1}{2}gt^{2} \Rightarrow 1 = 0 \times t_{1} + \frac{1}{2}gt_{1}^{2} \Rightarrow t_{1} = \sqrt{2/g}$
Velocity after travelling $1\ m$ distance
$v^{2} = u^{2} + 2gh \Rightarrow v^{2} = (0)^{2} + 2g \times 1 \Rightarrow v = \sqrt{2g}$
For second $1\ m$ distance
$1 = \sqrt{2g} \times t_{2} + \frac{1}{2}gt_{2}^{2} \Rightarrow gt_{2}^{2} + 2\sqrt{2g}t_{2} - 2 = 0$
$t_{2} = \frac{- 2\sqrt{2g} \pm \sqrt{8g + 8g}}{2g} = \frac{- \sqrt{2} \pm 2}{\sqrt{g}}$
Taking $+ ve$ sign $t_{2} = (2 - \sqrt{2})/\sqrt{g}$
$\therefore\frac{t_{1}}{t_{2}} = \frac{\sqrt{2/g}}{(2 - \sqrt{2})/\sqrt{g}} = \frac{1}{\sqrt{2} - 1}$ and so on
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.