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Q13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half 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 and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10 m s ^{-1}$ ?

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The volume of the drop is :

$V\ =\ \frac{4}{3}\pi r^3\ =\ \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3$

Thus the mass of raindrop is :

$m\ =\ \rho v$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ Kg$

Thus the work done is given by :

$W\ =\ F.s$

or $=\ mgs$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3 \times 9.8\ \times 250$

or $=\ 0.082\ J$

Now the total energy at the peak point is :

$E_p\ =\ mgh\ +\ 0\ =\ mgh$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times 9.8 \times 500$

or $=\ 0.146\ J$

And the energy at the ground is :

$E_b\ =\ 0\ +\ \frac{1}{2}mv^2\ =\ \frac{1}{2}mv^2$

or $=\ \frac{1}{2}\times 10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times (10)^2$

or $=\ 1.67\times 10^{-3} \ J$

Thus work done by the resistive force is :

$=\ 1.67\times 10^{-3} \ J\ -\ 0.164\ J\ =\ -\ 0.162\ J$

Posted by

Devendra Khairwa

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