A) 6
B) 10
C) 12
D) 4
Correct Answer: B
Solution :
Key Idea Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement. By definition \[{{V}_{i}}>0\] ie, \[{{V}_{0}}>{{V}_{i}}\] So, if in time t the angular speed of a body changes from \[{{\omega }^{2}}\] to \[{{\omega }^{2}}\] \[2m{{l}^{2}}\] If \[\sqrt{3}m{{l}^{2}}\] is constant \[3m{{l}^{2}}\] or \[m{{l}^{2}}\] ?.(i) Now, as by definition \[{{T}_{0}}\] Eq. (i) becomes \[{{T}_{0}}\] ie, \[{{T}_{f}}\] So, if in the time t angular displacement is \[{{T}_{0}}\]. \[{{T}_{f}}=\frac{3}{7}{{T}_{0}}\] or \[{{T}_{f}}=\frac{7}{3}{{T}_{0}}\] ?....(ii) Given, \[{{T}_{f}}=\frac{3}{2}{{T}_{0}}\] Hence, \[{{T}_{f}}=\frac{5}{2}{{T}_{0}}\] or \[\sqrt{2}qa\] Note Eqs. (i) and (ii) are similar to first and second equations of linear motion.
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