A) 3.2 m/s
B) 5.4 m/s
C) 7.6 m/s
D) 9.2 m/s
Correct Answer: B
Solution :
In this process potential energy of the metre stick will be converted into rotational kinetic energy. P.E. of metre stick\[=mg\left( \frac{\ell }{2} \right)\] \[\because \]Its centre of gravity lies at the middle point of the rod. Rotational kinetic energy \[E=\frac{1}{2}I{{\omega }^{2}}\] Here \[I=\frac{m{{\ell }^{2}}}{3}\] \[\therefore \] By the law of conservation of energy, \[mg\left( \frac{\ell }{2} \right)=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}.\frac{m{{\ell }^{2}}}{3}\left( \frac{{{v}_{n}}}{\ell } \right)\] \[\therefore \]By solving we get \[{{v}_{B}}=\sqrt{3g\ell }=5.4\,m/s.\] Hence, the correction option is [b].
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