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Question: Consider a drop of rainwater having mass \(1g\) falling from a height of \(1km\). It hits the ground...

Consider a drop of rainwater having mass 1g1g falling from a height of 1km1km. It hits the ground with a speed of 50m/s50m/s. Take 'gg' constant with a value 10m/s210m/{s^2}. The work is done by the (i)\left( i \right) gravitational force (ii)\left( {ii} \right) the resistive force of airs is:
(a)\left( a \right) (i)\left( i \right) 10J10J (ii)\left( {ii} \right) 8.75J - 8.75J
(b)\left( b \right) (i)\left( i \right) 10J10J (ii)\left( {ii} \right) 8.75J8.75J
(c)\left( c \right) (i)\left( i \right) 1.25J1.25J (ii)\left( {ii} \right) 8.75J - 8.75J
(d)\left( d \right) (i)\left( i \right) 100J100J (ii)\left( {ii} \right) 8.75J - 8.75J

Explanation

Solution

Hint First of all for this question we will calculate the work done by the gravitational force by using the formula Wg=mghWg = mgh and then we will find out the change in the kinetic energy and when we get these two values, by applying the work-energy theorem we will get the resistive force of the air.
Formula used:
Work done by the gravitational force will be given by
Wg=mghWg = mgh
Here,
WgWg, will be the work done duo gravitational force
mm, will be the mass
gg, will be the acceleration due to gravity
hh, will be the height

Complete Step By Step Solution
Firstly we will see the values which are given to us.
m=1g=11000kgm = 1g = \dfrac{1}{{1000kg}}
h=1km=1000mh = 1km = 1000m
Presently by applying the equation
Work done by the gravitational power will be given by
Wg=mghWg = mgh
Substituting the values, we get
Wg=11000×10×1000\Rightarrow Wg = \dfrac{1}{{1000}} \times 10 \times 1000
On understanding the above condition, we get
Wg=10J\Rightarrow Wg = 10J
Therefore, the10J10J work is done by the gravitational force.
Now we will see the change in the kinetic energy
And it will be given by
K.E=12mv20\vartriangle K.E = \dfrac{1}{2}m{v^2} - 0
Now we will substitute the values, we get
1211000×50×50\Rightarrow \dfrac{1}{2}\dfrac{1}{{1000}} \times 50 \times 50
Therefore on solving the above equation, we get
1.25J\Rightarrow 1.25J
Now by using the work-energy theorem,
We know
Work energy theorem=Wg+Wair=K.E = {W_g} + {W_{air}} = \vartriangle K.E
So now on substituting the values, we get
10J+resistance=1.2510\Rightarrow 10J + res{\text{istance}} = 1.25 - 10
From here we will calculate the value of resistance,
And it will be equal to
Resistance=8.75J = - 8.75J

Hence the option AA is correct.

Note The work-energy hypothesis expresses that the work done by all powers following up on a molecule approaches the adjustment in the molecule's active energy. The work-energy guideline expresses that the difference in energy of a framework is equivalent to work done on or by the framework. I don't generally know a thorough confirmation however this standard depends on the law of preservation of energy, which is one the key and significant laws that are utilized wherever in material science. Energy when any cooperation, impact, and so on are consistently the same.