Question

A particle is dropped from rest from a large height. Assume $g$ to be constant throughout the motion. The time taken by it to fall through successive distances of $1 \,m$ each will 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 reciprocals of the square roots of the integers, i.e., $\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}},...$

Answer 1

Answer: (C)

Time taken to cover first $n\, m$ is given by
$n = \frac{1}{2} gt_n^2$ or $t_n = \sqrt{\frac{2n}{g}}$
Time taken to cover first $(n + 1) m$ is given by
$t_{n+1} = \sqrt{\frac{2(n+1)}{g}}$
So time taken to cover $(n + 1)^{th} m$ is given by
$t_{n+1} - t_n = \sqrt{\frac{2(n+1)}{g}} - \sqrt{\frac{2n}{g}}$
$=\sqrt{\frac{2}{g}} [\sqrt{n + 1} - \sqrt{n} ]$
This gives the required ratio as:
$\sqrt{1},\left(\sqrt{2} -\sqrt{1}\right),\left(\sqrt{3}-\sqrt{2}\right), ...$ etc
(starting from $n = 0$)