question_answer

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.  

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) \[1:\sqrt{2}:\sqrt{3}-1\]

Correct Answer: B

Solution :

\[\text{S=AB=}\frac{\text{1}}{\text{2}}\text{g}{{\text{t}}_{\text{1}}}^{\text{2}}\] \[\Rightarrow \text{2S=AC=}\frac{\text{1}}{\text{2}}\text{g}{{\left( {{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}} \right)}^{\text{2}}}\] \[\text{and 3S = AD = }\frac{1}{2}G{{\left( {{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}+{{t}_{3}} \right)}^{\text{2}}}\] \[{{t}_{1}}=\sqrt{\frac{2S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}=\sqrt{\frac{4S}{g}},{{t}_{2}}=\sqrt{\frac{4S}{g}}-\sqrt{\frac{2S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}+{{t}_{3}}=\sqrt{\frac{6S}{g}},{{t}_{3}}=\sqrt{\frac{6S}{g}}-\sqrt{\frac{4S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{:}{{\text{t}}_{\text{2}}}\text{:}{{\text{t}}_{\text{3}}}::1:\left( \sqrt{2}-\sqrt{1} \right):\left( \sqrt{3}-\sqrt{2} \right)\]