A) 10.8 m/s
B) 5.4 m/s
C) 2.7 m/s
D) None of these
Correct Answer: B
Solution :
Here, the loss in potential energy is equal to the gain in the rotational kinetic energy. Hence \[\therefore \,\,\,\,\operatorname{mhg}\,\,=\, \frac{1}{2}\,l{{\omega }^{2}}\] \[\operatorname{m} \left( 9.8 \right) \left( 0.5 \right) = \frac{1}{2}\,\,\left( \frac{m}{3} \right)\,{{\omega }^{2}}~~~~~[as r= 1 m]\] Or \[\omega =\sqrt{(3\times 9.8)}\,\,=\,\,\sqrt{(29.4)}\,\,=\,\,5.4\] Further, \[\operatorname{v} =r\omega =1 \times 5.4 = 5.4 = 5.4 m/s\]
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