A) \[4.2\text{ }m/s\]
B) \[1.4\text{ }m/s\]
C) \[2.8\text{ }m/s\]
D) \[\sqrt{6}\text{ }m/s\]
Correct Answer: A
Solution :
The centre of mass of the stick fall through\[0.3\text{ }m\]. According to law of conservation of energy \[\frac{1}{2}I{{\omega }^{2}}=mgh\] \[\Rightarrow \]\[\frac{1}{2}\,\frac{m{{\ell }^{2}}}{3}\,\frac{{{v}^{2}}}{{{\ell }^{2}}}=mgh\] \[(\because \,v=\omega \ell )\] Here \[h=\ell /2=0.3\,m\] \[V=\sqrt{6gh}=\sqrt{6\times 9.8\times 0.3}=4.2\,m/s\]
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