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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A particle is dropped vertically from rest from a certain height. Time taken by it to fall through successive distance of one metre each will be :

$\begin{array}{1 1}(A)\;\sqrt 1 : \sqrt 2 : \sqrt 3: \sqrt 4\\(B)\;1:1:1:1 \\(C)\;(\sqrt 1-0) :(\sqrt 2-\sqrt1) :(\sqrt 3 -\sqrt 3).... \\(D)\;\frac{1}{\sqrt 1}:\frac{1}{\sqrt 2} :\frac{1}{\sqrt 3} :\frac{1}{\sqrt 4}\end{array} $

Can you answer this question?
 
 

1 Answer

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$1= \large\frac{1}{2} $$gt_1^2 => t_1 = \bigg(\sqrt{ \large\frac{2}{g}}\bigg)$
$2= \large\frac{1}{2} $$g(t_1+t_2)^2 => t_1+t_2 = \bigg(\sqrt{ \large\frac{2}{g}}\bigg)$$\sqrt 2$
$2= \large\frac{1}{2}$$g (t_1+t_2+t_3)^2 => t_1+t_2+t_3=\bigg(\sqrt{ \large\frac{2}{g}}\bigg) $$ \sqrt 3$
$t_1 :t_2:t_3=\sqrt 1 : (\sqrt 2 -\sqrt 1 ): (\sqrt 3 - \sqrt 2)$
Hence C is the correct answer.
answered Jul 18, 2014 by meena.p
 

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