Consider a drop of rain water having mass 1g falling from a height of 1km. It hits the ground with a speed of 50m/s. Take 'g' constant with a value 10m/s2. The work done by the (i) gravitational force the (ii) resistive force of airs is:
(i)10J(ii)−8.75J
(i)10J(ii)8.75J
(i)100J(ii)−8.75J
(i)1.25J(ii)−8.75J
A
(i)100J(ii)−8.75J
B
(i)10J(ii)8.75J
C
(i)1.25J(ii)−8.75J
D
(i)10J(ii)−8.75J
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Solution
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The correct option is A(i)10J(ii)−8.75J m=1g=11000kg
h=1km=1000m
work done by gravitional force wg−mgh wg=11000×10×1000 Change in kinetic energy △K.E=12mv2−0 =1211000×50×50−0 =1.25J Work energy theorem: wg+wairresistance=△K.E 10J+resistance=1.25−10 where resistance=−8.75J
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