A) 2.5
B) 3.0
C) 3.5
D) 4.0
Correct Answer: A
Solution :
[a] : As no external torque is applied to the system, the angular momentum of the system remains conserved. \[\therefore \]\[{{L}_{i}}={{L}_{f}}\]where the subscripts represent initial and final. or \[{{I}_{i}}{{\omega }_{i}}={{I}_{f}}{{\omega }_{f}}\]. Substituting the given values, we get \[\therefore \]\[1\times 100=2\times 1{{\omega }_{f}}\] \[{{\omega }_{f}}=50\]rad\[{{\text{s}}^{-1}}\] ...(i) Initial kinetic energy,\[{{K}_{i}}=\frac{1}{2}{{I}_{i}}\omega _{i}^{2}\] \[=\frac{1}{2}\times 1\times {{(100)}^{2}}=5\times {{10}^{3}}J\] Final kinetic energy, \[{{K}_{f}}=\frac{1}{2}\times {{I}_{f}}\times \omega _{f}^{2}\] \[=\frac{1}{2}\times 2\times {{(50)}^{2}}\] (Using (i)) \[=2.5\times {{10}^{2}}J\] Loss in kinetic energy, \[\Delta K={{K}_{i}}-{{K}_{f}}\] \[=5\times {{10}^{3}}-2.5\times {{10}^{3}}J=2.5\times {{10}^{3}}J=2.5\text{ }kJ\]
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