A) \[\operatorname{increase} by 10%\]
B) \[\operatorname{decrease} by 10%\]
C) \[\operatorname{increase} by 20%\]
D) \[\operatorname{decrease} by 20 %\]
Correct Answer: B
Solution :
As, \[K=\frac{1}{2}m{{v}^{2}}\,\,\,\Rightarrow \,\,\,v=\sqrt{\frac{2K}{m}}\] \[\therefore \,\,\,\,\,\,v'=\sqrt{\frac{2K'}{m}}=\sqrt{\frac{2\times 0.81\,\,K}{m}}\,\,=\,\,0.9\,\,v\] \[\therefore \,\,\,%\] decrease in speed = \[=\,\,\frac{v-0.9\,v}{v}\times 100=10\,%\]
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