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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