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Question

Consider a drop of rain water having mass $1 g$ falling from a height of $1 k m$ . It hits the ground with a speed of $50 m / s$ . Take ' $g$ ' constant with a value $10 m / s^{2}$ . The work done by the (i) gravitational force the (ii) resistive force of airs is:

A

$(i) 10 J (i i) - 8.75 J$

B

$(i) 10 J (i i) 8.75 J$

C

$(i) 1.25 J (i i) - 8.75 J$

D

$(i) 100 J (i i) - 8.75 J$

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Updated on : 2022-09-05

Solution

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Correct option is A)

$m = 1 g = \frac{1}{1 0 0 0} k g$
$h = 1 k m = 1000 m$
work done by gravitional force $w_{g} - m g h$
$w g = \frac{1}{1 0 0 0} \times 10 \times 1000$
Change in kinetic energy $△ K . E = \frac{1}{2} m v^{2} - 0$
$= \frac{1}{2} \frac{1}{1 0 0 0} \times 50 \times 50 - 0$
$= 1.25 J$
Work energy theorem: $w_{g} + w_{a i r r e s i s t a n c e} = △ K . E$
$10 J + r e s i s t a n c e = 1.25 - 10$
where resistance $= - 8.75 J$

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