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Question

A dancer spins about a vertical axis at $$60$$$$\mathrm { rpm }$$ with her arms folded.If she streches her hands so that M.I. about the vertical axis increases by $$25 \% ,$$ the new rate of revolution is


A
48 rpm
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B
75 rpm
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C
15 rpm
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D
24 rpm
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Solution

The correct option is A $$48$$ $$rpm$$
$$\begin{array}{l}\text { Let, the initial moment of inertia of dancer }=I_{0} . \\\text { and, initial rate of revolution }=60\mathrm{rpm}\text { . } \\\text { Given, final moment of inertia }=I_{0}+I_{0} \times \frac{25}{100}=\frac{5}{4} \text { Io. }\end{array}$$

$$\begin{array}{l}\text { Here, there is no any external torque, so angular } \\\text { momentum remains conserved. }\end{array}$$


$$\begin{array}{r}\text { Thus, } I_{i}\omega_{i}=I_{f}\omega_{f}\Rightarrow I_{0} \times 60=\frac{5}{4} I_{0} \times \omega_{F} \\\text { Thus, }\omega_{F}=48\mathrm{rpm}\end{array}$$

Physics

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