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A rain drop of radius $2 \mathrm{~mm}$ falls from a height of 500 $\mathrm{m}$ above the ground. It falls with decreasing acceleration due to viscous resistance of air until half of its original height. It attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first half and second half of the journey? [Density of $\mathrm{H}_2 \mathrm{O}=10^3 \mathrm{~kg}$ / $\mathrm{m}^3$. What is the work done by the resistance force in the entire journey if its speed on reaching the ground is $10 \mathrm{~m} / \mathbf{s}$ ?
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Here, $r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}, \rho=10^3 \mathrm{~kg} / \mathrm{m}^3$
Distance moved in half of the journey $=\mathrm{S}=\frac{500}{2}=250 \mathrm{~m}$.
Mass of the drop $=$ Vol. $\times$ density $=\frac{4}{3} \pi r^3 \rho$
$$
=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3=3.35 \times 10^{-5} \mathrm{~kg}
$$
Work done $=\mathrm{W}=\mathrm{F} \times \mathrm{S}=3.35 \times 10^{-5} \times 9.8 \times 250=0.082 \mathrm{~J}$
Work done by the gravitational force remains same weather the drop falls under decreasing acceleration or with uniform speed. So, the work done in second half of the journey is also $0.082 \mathrm{~J}$.
Energy of the drop on reaching the ground $=E=m g h=3.35 \times 10^{-5} \times 9.8 \times 500=0.164 \mathrm{~J}$ (considering no resistive forces).
Kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 3.35 \times 10^{-5} \times 10^2=1.675 \times$ $10^{-3} \mathrm{~J}=0.001675=0.1623 \mathrm{~J}$
$=$ Work done by the resistive force.
Distance moved in half of the journey $=\mathrm{S}=\frac{500}{2}=250 \mathrm{~m}$.
Mass of the drop $=$ Vol. $\times$ density $=\frac{4}{3} \pi r^3 \rho$
$$
=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3=3.35 \times 10^{-5} \mathrm{~kg}
$$
Work done $=\mathrm{W}=\mathrm{F} \times \mathrm{S}=3.35 \times 10^{-5} \times 9.8 \times 250=0.082 \mathrm{~J}$
Work done by the gravitational force remains same weather the drop falls under decreasing acceleration or with uniform speed. So, the work done in second half of the journey is also $0.082 \mathrm{~J}$.
Energy of the drop on reaching the ground $=E=m g h=3.35 \times 10^{-5} \times 9.8 \times 500=0.164 \mathrm{~J}$ (considering no resistive forces).
Kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 3.35 \times 10^{-5} \times 10^2=1.675 \times$ $10^{-3} \mathrm{~J}=0.001675=0.1623 \mathrm{~J}$
$=$ Work done by the resistive force.
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