Question

A particle is dropped from rest from a height. The time taken by it to fall through successive vertical distance 1 meter each will then be :-

• A. all equal, being equal to $\frac{2}{g}$ second
• B. in the ratio of square roots of integers $1,2,3,\dots$
• C. in the ratio of the difference in the square roots of integers $(\sqrt{1}-\sqrt{0}),(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})$
• D. in the ratio $\frac{1}{\sqrt{1}}:\frac{1}{\sqrt{2}}:\frac{1}{\sqrt{3}}:\frac{1}{\sqrt{4}}$

Answer 1

Answer: (C)

For first $1m$ of fall,
$1=\frac{1}{2}g{t}_1^2,$
$\therefore t_1=\sqrt{\frac{2}{g}}$
For $2m$ of fall,
$2=\frac{1}{2}g{t}^2,$
$\therefore t=\sqrt{\frac{4}{g}},$
$\therefore t_2=t-t_1=\sqrt{\frac{4}{g}}-\sqrt{\frac{2}{g}}=(\sqrt{2}-1)\sqrt{\frac{2}{g}}$
For $3m$ of fall,
$3=\frac{1}{2}g{t}^2,$
$\therefore t=\sqrt{3}\sqrt{\frac{2}{g}},$
$\therefore t_3=t-t_2=(\sqrt{3}-\sqrt{2})\sqrt{\frac{2}{g}}$