| An automatic rotating water sprinkler is mounted over top of a 5 m high pillar in a field. |
 |
| Water is flowing at a rate of \[5\times {{0}^{-\,4}}{{m}^{3}}\]in 1 s through a nozzle of area\[1\text{ }c{{m}^{2}}\]. Maximum possible area of field that can be irrigated by the nozzle is (take, \[g=10m{{s}^{-\,2}}\]) |
A) \[15\pi \,{{m}^{2}}\]
B) \[25\pi \,{{m}^{2}}\]
C) \[100\pi \,{{m}^{2}}\]
D) \[75\pi \,{{m}^{2}}\]
Correct Answer: B
Solution :
| \[{{v}_{\text{outflow}}}=\frac{\text{Volume}\,\text{flow}\,\text{rate}}{\text{Area}}\] |
| \[=\frac{5\times {{10}^{-\,4}}}{1\times {{10}^{-\,4}}}=5\,m{{s}^{-\,1}}\] |
| For vertical motion of spray, |
| \[{{u}_{y}}=u\cdot \sin \theta =5\times \frac{1}{\sqrt{2}}\] |
| [For range mariinum, \[\theta =45{}^\circ \]] |
| \[h=-5\,m\] |
| Using \[h=ut+\frac{1}{2}a{{t}^{2}},\] we have |
| \[\Rightarrow \]\[-\,5=\frac{5}{\sqrt{2}}\cdot t-\frac{1}{2}\times 10\times {{t}^{2}}\] |
| \[\Rightarrow \]\[5{{t}^{2}}-\frac{5}{\sqrt{2}}t-5=0\] |
| \[{{t}^{2}}-\frac{t}{\sqrt{2}}-1=0\] |
| Applying Sridharacharya rule, we get \[\Rightarrow \]\[t=\frac{\frac{1}{\sqrt{2}}\pm \sqrt{\frac{1}{2}-4(1)(-1)}}{2\times 1}=\frac{\frac{1}{\sqrt{2}}\pm \frac{3}{\sqrt{2}}}{2}\] |
| \[=\frac{4}{2\sqrt{2}}=\sqrt{2}s\] |
| During this time water keeps on moving in forward horizontal direction with speed \[u\cos \theta .\] |
| So, water spray can cover a distance is \[u\cos \theta \times t.\] |
| Radius of circular region of spray \[=5\times \frac{1}{\sqrt{2}}\times \sqrt{2}=5\,m\] |
| Hence, Area irrigated \[=\pi {{R}^{2}}=25\pi {{m}^{2}}\] |
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