Q172 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
An object is dropped from a certain point A at a height 'h' from the ground. During it's journey straight downwards, the object passes points B and C such that the ratio of time taken $t_1$ to cover AB and $t_2$ to cover BC is 1: $(\sqrt{2} - 1)$. What is the ratio of distances AB:BC ?
- A$1:\sqrt{2}$
- B$\sqrt{2}:1$
- Correct answerC$1:1$
- D$(\sqrt{2} - 1):1$
Solution
Let the time taken to travel from A to B be $t_1 = t$.
The time taken to travel from B to C is $t_2 = (\sqrt{2} - 1)t$.
The total time taken to travel from A to C is $t_{AC} = t_1 + t_2 = t + (\sqrt{2} - 1)t = \sqrt{2}t$.
Since the object is dropped from rest at A, the initial velocity is zero.
The distance AB is given by $AB = \dfrac{1}{2}gt_1^2 = \dfrac{1}{2}gt^2$.
The distance AC is given by $AC = \dfrac{1}{2}gt_{AC}^2 = \dfrac{1}{2}g(\sqrt{2}t)^2 = gt^2$.
The distance BC is $BC = AC - AB = gt^2 - \dfrac{1}{2}gt^2 = \dfrac{1}{2}gt^2$.
The ratio of distances AB and BC is $AB : BC = \dfrac{1}{2}gt^2 : \dfrac{1}{2}gt^2 = 1 : 1$.
Answer: $1:1$
The time taken to travel from B to C is $t_2 = (\sqrt{2} - 1)t$.
The total time taken to travel from A to C is $t_{AC} = t_1 + t_2 = t + (\sqrt{2} - 1)t = \sqrt{2}t$.
Since the object is dropped from rest at A, the initial velocity is zero.
The distance AB is given by $AB = \dfrac{1}{2}gt_1^2 = \dfrac{1}{2}gt^2$.
The distance AC is given by $AC = \dfrac{1}{2}gt_{AC}^2 = \dfrac{1}{2}g(\sqrt{2}t)^2 = gt^2$.
The distance BC is $BC = AC - AB = gt^2 - \dfrac{1}{2}gt^2 = \dfrac{1}{2}gt^2$.
The ratio of distances AB and BC is $AB : BC = \dfrac{1}{2}gt^2 : \dfrac{1}{2}gt^2 = 1 : 1$.
Answer: $1:1$
Hence, the correct answer is option: 'C'.