A) 14m
B) 13.11m
C) 10m
D) 2.0m
Correct Answer: B
Solution :
Let A be vector in\[x-y\]plane. Its\[x\]and y components are \[{{A}_{x}}=12\text{ }m\] and \[{{A}_{y}}=8\text{ }m\] \[A=\sqrt{A_{x}^{2}+A_{y}^{2}}\] \[=\sqrt{{{(12)}^{2}}+{{(8)}^{2}}}\] \[A=\sqrt{208}\,m\] When the vector is rotated in\[x-y\]plane, then \[x\]component become halved and its new y component \[A_{y}^{}=\sqrt{{{\left( \frac{{{A}_{x}}}{2} \right)}^{2}}+A_{y}^{2}}\] \[\sqrt{208}=\sqrt{{{(6)}^{2}}+A_{y}^{2}}\] \[A_{y}^{}=\sqrt{208-36}\] \[A_{y}^{}=\sqrt{172}\] \[=13.11cm\]
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