A) 19%
B) 38%
C) 57%
D) 76%
Correct Answer: B
Solution :
In first case,\[{{W}_{1}}=\frac{1}{2}m{{(12)}^{2}}+mg\times 12\] \[=\frac{1}{2}m\times 144+m\times 10\times 12\] \[=192\,\,m\] In second case, \[{{W}_{2}}=mgh=\mathbf{120}\,\,\mathbf{m}\] \[\therefore \]\[%\]energy saved\[=\frac{192m-120m}{192m}\times 100%\] \[=\mathbf{38%}\]You need to login to perform this action.
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