A) \[{{\omega }_{1}}>{{\omega }_{2}}\]
B) \[{{\omega }_{1}}={{\omega }_{2}}\]
C) \[{{\omega }_{2}}>{{\omega }_{1}}\]
D) None of these
Correct Answer: C
Solution :
Consider \[{{I}_{1}}\] and \[{{I}_{2}}\] be the moments of inertia of the hollow cylinder and solid sphere about its axis through its centre respectively. Then, \[{{I}_{1}}=M{{R}^{2}}\] ... (i) and \[{{I}_{2}}=\frac{2}{5}M{{R}^{2}}\] ... (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)
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