Import Question JSON

$\text{Hint: } h = ut + \frac{1}{2}gt^2$ $\text{Step 1: Find the time taken by the particle through successive distances of 1 m.}$ $h = ut + \frac{1}{2}gt^2 \Rightarrow 1 = 0 \times t_1 + \frac{1}{2}gt_1^2 \Rightarrow t_1 = \sqrt{\frac{2}{g}}$ $\text{Velocity after travelling 1m distance}$ $v^2 = u^2 + 2gh \Rightarrow v^2 = (0)^2 + 2g \times 1 \Rightarrow v = \sqrt{2g}$ $\text{For the second 1 meter distance}$ $1 = \sqrt{2g} \times t_2 + \frac{1}{2}gt_2^2 \Rightarrow gt_2^2 + 2\sqrt{2g}t_2 - 2 = 0$ $t_2 = \frac{-2\sqrt{2g} + \sqrt{8g + 8g}}{2g} = \frac{-\sqrt{2} + 2}{\sqrt{g}}$ $\text{Step 2: Find the ratio of the time.}$ $\text{Taking +ve sign } t_2 = \frac{(2 - \sqrt{2})}{\sqrt{g}}$ $\therefore \frac{t_1}{t_2} = \frac{\sqrt{2}/g}{(2 - \sqrt{2})/\sqrt{g}} = \frac{1}{\sqrt{2} - 1} \text{ and so on.}$ $\text{Therefore, the ratio of the difference in the square roots of the integers is } \sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), (\sqrt{4} - \sqrt{3}) \ldots$ $\text{Hence, option (3) is the correct answer.}$