A ball is released from the top of a tower, The ratio of work done by force of gravity in first, second and third second of the motion of ball is

1:2:3

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1:4:16

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1:3:5

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1:9:25

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Solution

The correct option is C 1:3:5
$W o r k d o n e = f o r c e \times d i s t a n c e$
$F o r c e = m a s s \times a c c e l e r a t i o n$
$∴ W o r k d o n e = m g h$
$m g h_{1} : m g h_{2} : m g h_{3} = h_{1} : h_{2} : h_{3}$
$= g (1 - \frac{1}{2}) : g (2 - \frac{1}{2}) : g (3 - \frac{1}{2})$
Height covered in nth second, $h_{n} = u + g (n - \frac{1}{2}) a n d u = 0$
$∴ R a t i o i s 1 : 3 : 5$

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A ball is released from the top of a tower, The ratio of work done by force of gravity in first, second and third second of the motion of ball is

The correct option is C 1:3:5Work done=force×distance Force=mass×acceleration ∴Work done=mgh mgh1:mgh2:mgh3=h1:h2:h3 =g(1−12):g(2−12):g(3−12) Height covered in nth second, hn=u+g(n−12) and u=0 ∴ Ratio is 1:3:5