Question

A particle is dropped from rest 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., $\sqrt{1}(\sqrt{2}-\sqrt{1})(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}),.....$
• 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, i.e., $\sqrt{1}(\sqrt{2}-\sqrt{1})(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}),.....$

Answer 2

For first $1 m$ $1=\frac{1}{2} \times g \times t ^{2}$ $t _{1}=\sqrt{\frac{2}{ g }}$ For first $1 m$ $2=\frac{1}{2} \times g \times t ^{2}$ $t _{1}=\sqrt{2} \times \sqrt{\frac{2}{ g }}$ for second $1 m$ $t _{2}= t - t _{1}$ $=\sqrt{\frac{2}{ g }}(\sqrt{2}-\sqrt{1})$ similarly, $t _{3}=\sqrt{\frac{2}{ g }}(\sqrt{3}-\sqrt{2})$