question_answer 51)

                  It is a common observation that rain clouds can be at about a kilometer altitude above the ground.                 (a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. \[(g=10\,\,m/{{s}^{2}})\]                 (b) A typical rain drop is about 4mm 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 exert on you.                 (e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.                 (Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced thought it)                

Answer:

                    (a) Here, \[u=0,\,\,h=\,1\,km=1000\,m,\]                 \[g=10\,\,m{{s}^{-2}}\]                 Using \[{{\upsilon }^{2}}-{{u}^{2}}=2gh\] or \[\upsilon _{{}}^{2}=2gh\]                 or \[\upsilon =\sqrt{2gh}=\sqrt{2\times 10\times 1000}\]                 \[=141.4\text{ }m{{s}^{-1}}\]                 \[=141.4\times \frac{18}{5}km/{{h}^{-1}}=509.04\text{ }km/{{h}^{-1}}\]                 (b) Volume of drop, \[V=\frac{4}{3\,}\pi \,{{r}^{3}}\]                 \[=\frac{4}{3\,}\times 3.14\times 8\times {{10}^{-9}}\]                 \[=3.35\times {{10}^{-8}}{{m}^{3}}\]                 Mass of drop.,                 \[m=V\times \rho =3.35\times {{10}^{-8}}\times ~{{10}^{3}}=3.35\times {{10}^{-5}}kg\]                 \ Momentum of drop,                 \[p=m\upsilon =3.35\times {{10}^{-5}}\times 141.4\]                 \[=4.74\times {{10}^{-3}}kg\text{ }m{{s}^{-1}}\]                         (c) Time taken to flatten the drop,                 \[t=\frac{\text{diameter}\,\,\text{of}\,\,\text{drop}}{\text{speed}\,\,\text{of}\,\,\text{drop}}\]                 \[=\frac{4\times {{10}^{-3}}m}{141.4\,\,m{{s}^{-1\,\,}}}\,=\,\,28\,\,\times \,\,{{10}^{-6}}s\]                 (d) \[\text{Force}\,\,\text{=}\,\,\frac{\text{Change}\,\,\text{in}\,\,\text{momentum}}{\text{Time taken}}\]                 \[=\frac{(4.74\,\,\times \,{{10}^{-3}}\,-\,0)}{28\,\,\times \,\,{{10}^{-6}}\,}\] (c) Area of umbrella \[=\pi {{r}^{2}}=3.14\times {{(0.5)}^{2}}=0.785\,{{m}^{2}}\]                 Number of drops in 5 cm separation   \[\text{=}\frac{\,\text{Area}\,\,\text{of}\,\,\text{umbrella}}{{{\text{(}\,\,\text{distance}\,\,\text{of}\,\text{separation)}}^{\text{2}}}}\]                 \[=\frac{0.785{{m}^{2}}}{25\,\,c{{m}^{2}}}\,\,=\,\frac{0.785\,{{m}^{2}}}{25\,\,\times \,\,{{10}^{-4\,}}{{m}^{2}}}\,\]             = 314                 Force exerted by one drop = 169.3 N                 \ Force exerted by 314 drops \[=169.3\times 314\]             = 53160.2 N                 \[5.32\times {{10}^{4}}N\]