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Q.

A body freely falls a certain height onto the ground in a time $z .$ During the first one third of the interval it gains a kinetic energy $Δ K_{1}$ and during the last one third of the interval, it gains a kinetic energy $Δ K_{2} .$ the ratio $Δ K_{1} : Δ K_{2}$ is

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a

$4 : 1$

b

$5 : 1$

c

$1 : 5$

d

$1 : 4$

answer is C.

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Detailed Solution

As we know that
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$h = u t + \frac{1}{2} g t^{2} h = \frac{1}{2} g t^{2}$
First one third of interval, last one third of interval
$\Rightarrow Δ K_{1} \propto {(\frac{t}{3})}^{2}, Δ K_{2} \propto t^{2} - {(\frac{2 t}{3})}^{2} ∴ \frac{Δ K_{1}}{Δ K_{2}} = \frac{{(\frac{t}{3})}^{2}}{t^{2} - {(\frac{2 t}{3})}^{2}} = \frac{\frac{1}{9}}{1 - \frac{4}{9}} = \frac{1}{9} \times \frac{9}{5} = \frac{1}{5}$

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