A) 5 : 4
B) 4 : 5
C) 1 : 4
D) 4 : 1
Correct Answer: C
Solution :
\[E=mg\times \frac{4h}{5}\] \[P.E\,\,=\,\,mgh\] \[K.E=mgh-m{{g}^{2}}h/5=mgh/5\] \[=\,\,\frac{\frac{1}{5}mgh\times 5}{4h\,\times \,mg}\,\,=\,\,\frac{1}{4}\]You need to login to perform this action.
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