__Question 6.13__:^{–1}?

__Solution__:Radius of the rain drop,

*r*= 2 mm = 2 × 10

^{–3}m

^{3}

= (4/3) X 3.14 X (2 X 10

^{-3})

^{3}m

^{-3}

Density of water,

*ρ*= 10

^{3}kg m

^{–3 }

Mass of the rain drop,

*m*=

*ρ*

*V*

= (4/3) X 3.14 X (2 X 10

^{-3})

^{3}X 10

^{3}kg

Gravitational force,

*F*=

*m*g

= (4/3) X 3.14 X (2 X 10

^{-3})

^{3}X 10

^{3}X 9.8 N

The work done by the gravitational force on the drop in the first half of its journey:

*W*

_{I}

*= Fs*

*=*(4/3) X 3.14 X (2 X 10

^{-3})

^{3}X 10

^{3}X 9.8 X 250 = 0.082 J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e.,

*W*

_{II}, = 0.082 J

As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.

∴Total energy at the top:

*E*

_{T}=

*mgh*+ 0

= (4/3) X 3.14 X (2 X 10

^{-3})

^{3}X 10

^{3}X 9.8 X 500 X 10

^{-5}

= 0.164 J

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.

∴Total energy at the ground:

E

_{G}= (1/2) mv

^{2}+ 0

= (1/2) X (4/3) X 3.14 X (2 X 10

^{-3})

^{3}X 10

^{3}X 9.8 X (10)

^{2}

^{}

= 1.675 X 10

^{-3}J

∴Resistive force =

*E*

_{G }–

*E*

_{T}= –0.162 J

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