A) \[{{\omega }_{1}}>{{\omega }_{2}}\]
B) \[{{\omega }_{1}}={{\omega }_{2}}\]
C) \[{{\omega }_{2}}>{{\omega }_{1}}\]
D) None of these
Correct Answer: C
Solution :
Consider \[{{\text{I}}_{1}}\] and \[{{\text{I}}_{2}}\] be the moments of inertia of the hollow cylinder and solid sphere about its axis through its centre respectively. Then, \[{{\text{I}}_{1}}=M{{R}^{2}}\] ? (i) and \[{{\text{I}}_{2}}=\frac{2}{5}M{{R}^{2}}\] ...(ii) Let \[\tau \] be the magnitude of the torque applied on each of them. If \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] are the angular accelerations produced in the cylinder and sphere respectively, then \[\tau ={{I}_{1}}{{\alpha }_{1}}\] and \[\tau ={{I}_{2}}{{\alpha }_{2}}\] \[\therefore \] \[{{I}_{1}}{{\alpha }_{1}}={{I}_{2}}{{\alpha }_{2}}\Rightarrow \frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{{{I}_{2}}}{{{I}_{1}}}\] \[=\frac{\frac{2}{5}M{{R}^{2}}}{M{{R}^{2}}}=\frac{2}{5}\Rightarrow {{\alpha }_{2}}=\frac{5}{2}{{\alpha }_{1}}\] \[\Rightarrow \] \[{{\alpha }_{2}}=2.5{{\alpha }_{1}}\] ? (iii) If \[{{\omega }_{1}}\] and \[{{\omega }_{2}}\] be the angular speed of the cylinder and sphere after time t, then \[{{\omega }_{1}}={{\omega }_{0}}+{{\alpha }_{1}}t\] ? (iv) and \[{{\omega }_{2}}={{\omega }_{0}}+{{\alpha }_{2}}t\] ? (v) where, \[{{\omega }_{0}}=\]initial angular speed \[\therefore \]From Eq. (iv) and (v), it is clear that \[{{\omega }_{2}}>{{\omega }_{1}}\] \[\therefore \]The sphere will acquire more angular speed as compared to that of the cylinder after a given time.
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