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                                    A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed `omega_0`. Now, the child folds his hands back so that moment of inertia becomes `3` times the initial value. The new angular speed is.A. `3omega_o`B. `omega_o`/3C. `6omega_o`D. `omega_o`/6 | 
                            
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Answer» Correct Answer - b Here, Initial angular speed, `omega_(i) = omega_(0)` Initial moment of inertia =`I_(i)` Final moment of inertia `I_(f)=3I_(i)` According to the law of conservation of angualr momentum, we get `L_(i)=L_(f)`or `I_(i)omega_(i)=I_(f)omega_(f)` `omega_(f) =(I_(i)omega_(i))/(I_(f)) = (I_(i))/(3I_(i))omega_(0)=omega_(0)/3`  | 
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