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 $1m$ 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 $nm$ is given by

$n=\frac{1}{2}g{t}_n^2$ 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 $n=0$)