Kinematics Question 403
Question: Let A, B, C, D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the times of descent through AB, BC and CD are in the ratio.
Options:
A) $ 1:\sqrt{3}-\sqrt{2}:\sqrt{3}+\sqrt{2} $
B) $ 1:\sqrt{2}-1:\sqrt{3}-\sqrt{2} $
C) $ 1:\sqrt{2}-1:\sqrt{3} $
D) $ 1:\sqrt{2}:\sqrt{3}-1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \text{S=AB=}\frac{1}{2}g{{t} _{1}}^{2} $
$ \Rightarrow \text{2S=AC=}\frac{1}{2}g{{( {{t} _{1}}\text{+}{{t} _{2}} )}^{2}} $
$ \text{and 3S = AD = }\frac{1}{2}G{{( {{t} _{1}}\text{+}{{t} _{2}}+t _{3} )}^{2}} $
$ t _{1}=\sqrt{\frac{2S}{g}} $
$ {{t} _{1}}\text{+}{{t} _{2}}=\sqrt{\frac{4S}{g}},t _{2}=\sqrt{\frac{4S}{g}}-\sqrt{\frac{2S}{g}} $
$ {{t} _{1}}\text{+}{{t} _{2}}+t _{3}=\sqrt{\frac{6S}{g}},t _{3}=\sqrt{\frac{6S}{g}}-\sqrt{\frac{4S}{g}} $
$ {{t} _{1}}\text{:}{{t} _{2}}\text{:}{{t} _{3}}::1:( \sqrt{2}-\sqrt{1} ):( \sqrt{3}-\sqrt{2} ) $
