A) \[\frac{16}{25}\]
B) \[\frac{2}{5}\]
C) \[\frac{3}{5}\]
D) \[\frac{9}{25}\]
Correct Answer: B
Solution :
| Key Idea: According to conservation of energy, potential energy at height h = kinetic energy at ground |
| Potential energy = Kinetic energy |
| i.e., \[mgh=\frac{1}{2}m{{v}^{2}}\] |
| \[\Rightarrow \] \[v=\sqrt{2gh}\] |
| If \[{{h}_{1}}\] and \[{{h}_{2}}\] are initial and final heights, then |
| \[{{v}_{1}}=\sqrt{2g{{h}_{1}}},\,{{v}_{2}}=\sqrt{2g{{h}_{2}}}\] |
| Loss in velocity |
| \[\Delta v={{v}_{1}}-{{v}_{2}}=\sqrt{2g{{h}_{1}}}-\sqrt{2g{{h}_{2}}}\] |
| \[\therefore \,\] Fractional loss in velocity \[=\frac{\Delta v}{{{v}_{1}}}\] |
| \[=\frac{\sqrt{2g{{h}_{1}}}-\sqrt{2g{{h}_{2}}}}{\sqrt{2g{{h}_{1}}}}\] |
| \[=1-\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}\] |
| Substituting the values, we have |
| \[\therefore \] \[\frac{\Delta v}{{{v}_{1}}}=1-\sqrt{\frac{1.8}{5}}\] |
| \[=1-\sqrt{0.36}=1-0.6\] |
| \[=0.4=\frac{2}{5}\] |
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