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 distance 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,
  $\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}),.....$
• 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)

In the ratio of the difference in the square roots of the integers,

$\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}),.....$
Time taken to cover first n metre 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
$\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),....$etc (starting from n = 0)