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It is a common observation that rain clouds can be at about a kilometre altitude above the ground.

a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h

b) A typical raindrop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.

c) Estimate the time required to flatten the drop.

d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.

e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two raindrops is 5 cm.

Answers (1)

$\\\; h=1\; km=1000\; m\\g=10\; m/s^{2}\\d=4\; mm\; and\; u=0\; m/s$

Thus, $\\r=\frac{4}{2}\; mm\\=2\times10^{-3}m$

(a) Let's find out the velocity of raindrop on the ground’

$\\v^{2}=u^{2}-2as\\=u^{2}-2g(-h)\\=u^{2}+2gh\\=0^{2}+2(10)(1000)$

Thus,

$\\v=100\sqrt{2}\; m/s\\v=100\sqrt{2}\left ( \frac{18}{5} \right )km/hr$

$\\=360\sqrt{2}\; km/hr\\=510\; km/hr$

(b) Momentum of the raindrop when it touches the ground

mass of drop(m) = Vol. × density

$=\frac{4}{3}\pi \; r^{3}\rho$ $(\rho = density\; of\; water)$

Now, density of water $=10^{3}kg/m^{3}$

Thus, $M=\frac{4}{3}\pi (2\times10^{-3})^{3}\times1000$

$=\frac{4}{3}\times\frac{22}{7}\times2\times2\times2\times10^{-9}\times10^{3}$

$=\frac{704}{21}\times10^{-6}kg$

$=3.35\times10^{-5}kg$

Now, we know that Momentum (p) = mv

$p=3.35\times10^{-5}\times100\sqrt{2}$

$=4.7\times10^{-3}kg\; ms^{-1}$

(c) Time required for a drop to be flattened-

$Time=\frac{distance}{speed}$

$=\frac{4\times10^{-3}}{100\sqrt{2}}m\; \; \; \; \; \; .....(distance=4mm=4\times10^{-3})$

$=4\frac{(\sqrt{2})}{100(2)}\times10^{-3}$

$=\frac{2(1.414)}{100}\times10^{-3}$

$=2.8\times10^{-5}\; sec$

(d) Now, we know that,

$Force=\frac{dp}{dt}$

$=\frac{mv-0}{t-0}$

$Force=\frac{4.7\times10^{-3}}{2.8\times10^{-5}}$

$=1.68(10^{2})$

$=168\; N$

(e) Here,

Radius of umbrella $=\frac{1}{2}m\; \; \; \; \; \; .......(since\; its\; \; diameter=1m)$

Thus, Area of umbrella $=\pi R^{2}$

$=\frac{22}{7}.\frac{1}{2}.\frac{1}{2}m^{2}$

$=\frac{11}{14}m^{2}$

Now, the square area covered by one drop

$= (5 \times 10^{-2})^{2}$

$= 25 \times 10^{-4} m^{2}$

Therefore, no. of drops falling on the umbrella $= \frac{\pi R^{2}}{25} \times 10^{-4}$

$=\frac{11(10)^{4}}{14(25)}$

$=0.0314\times10^{4}$

Therefore 314 drops fell on the umbrella.

Thus, the net force on the umbrella $= 314 \times 168N = 52752 N.$

Posted by

infoexpert22

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