We know that angular momentum of spin $=I \omega$ By the conservation of angular momentum $\frac{2}{5} M R^{2} \cdot \frac{2 \pi}{T}=\frac{2}{5} M\left(\frac{R}{4}\right)^{2} \cdot \frac{2 \pi}{T'}$ $T'=\frac{T}{16}=\frac{24}{16}=1.5\, h$
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