A) twice
B) thrice
C) four times
D) same
Correct Answer: B
Solution :
\[K{{E}_{1}}=\frac{1}{2}m\,(v_{2}^{2}-v_{1}^{2})\] \[\therefore \] \[K{{E}_{1}}=\frac{1}{2}m\,[{{(20)}^{2}}-{{(10)}^{2}}]\] \[\Rightarrow \] \[K{{E}_{1}}=\frac{1}{2}\,m\,(400-100)\] \[\Rightarrow \] \[K{{E}_{1}}=\frac{1}{2}m\times 300=150\,m\] ... (i) \[K{{E}_{2}}=\frac{1}{2}\,m\,[{{(10)}^{2}}-0]\] \[\Rightarrow \] \[K{{E}_{2}}=\frac{1}{2}\,m\times 100=50\,m\] ... (ii) From Eqs. (i) and (ii), we have \[K{{E}_{1}}=3K{{E}_{2}}\]
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