A) 5
B) \[5\sqrt{2}\]
C) \[5\sqrt{3}\]
D) \[10\sqrt{2}\]
Correct Answer: C
Solution :
It is given \[{{v}_{\max }}=100\,cm/sec\], a = 10 cm. Þ \[{{v}_{\max }}=a\omega \]Þ \[\omega =\frac{100}{10}=10\,rad/sec\] Hence \[v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\] Þ\[50=10\sqrt{{{(10)}^{2}}-{{y}^{2}}}\] Þ \[y=5\sqrt{3}\,cm\]
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