A) \[\frac{{{v}_{0}}}{2g{{t}_{0}}}\]
B) \[\frac{{{v}_{0}}}{4g{{t}_{0}}}\]
C) \[\frac{3{{v}_{0}}}{4g{{t}_{0}}}\]
D) \[\frac{{{v}_{0}}}{g{{t}_{0}}}\]
Correct Answer: A
Solution :
[A] \[\frac{3}{4}th\] energy is lost i.e.,\[\frac{1}{4}th\] kinetic energy is left. Hence, its velocity becomes \[\frac{{{v}_{0}}}{2}\] under a retardation of \[\mu g\] in time to. \[\therefore \,\,\frac{{{v}_{0}}}{2}={{v}_{0}}-\mu g\,\,{{t}_{0}}\] or \[\mu g\,\,{{t}_{0}}=\frac{{{v}_{0}}}{2}\] or \[\mu =\frac{{{v}_{0}}}{2g{{t}_{0}}}\]
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