A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1m each will then be

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All equal, being equal to

\sqrt{2 / g}

second

In the ratio of square roots of the integers 1, 2, 3, ….

In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), (\sqrt{4} - \sqrt{3}) .......$

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}}

answer is C.

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Detailed Solution

The time taken to cover 'n' meters is given by

n = $\frac{1}{2} g {t_{n}}^{2}$

$t_{n} = \sqrt{\frac{2 n}{g}}$

Time taken to cover (n+1) meters is given by

$t_{n + 1} = \sqrt{\frac{2 (n + 1)}{g}}$

Therefore, $t_{n + 1} - t_{n} = \sqrt{\frac{2 (n + 1)}{g}} - \sqrt{\frac{2 n}{g}}$

= $\sqrt{\frac{2}{g}} [\sqrt{n + 1} - \sqrt{n}]$

This gives $t h e r a t i o o f t h e d i f f e r e n c e i n t h e s q u a r e r o o t s o f t h e i n t e g e r s$ $\sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), (\sqrt{4} - \sqrt{3}) .......$