A wheel is rotating freely with an angular speed $\inline \omega$ on a shaft. The moment of inertia of the wheel is $\inline I$ and the moment of inertia of the shaft is negligible . Another wheel of moment of inertia $\inline 3I$ initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is : Option: 1 Option: 2 Option: 3 Option: 4

\begin{aligned} &\text { By anglar momentum conservation }\\ &\omega I+3 I \times 0=4 I \omega^{\prime} \Rightarrow \omega^{\prime}=\frac{\omega}{4}\\ &(\mathrm{KE})_{\mathrm{i}}=\frac{1}{2} \mathrm{I} \omega^{2}\\ &(\mathrm{KE})_{\mathrm{f}}=\frac{1}{2} \times(4 \mathrm{I}) \times\left(\frac{\omega}{4}\right)^{2}=\frac{\mathrm{I} \omega^{2}}{8}\\ &\Delta \mathrm{KE}=\frac{3}{8} \mathrm{I} \omega^{2}\\ &\text { fractional loss }=\frac{\Delta \mathrm{KE}}{\mathrm{KE}_{1}}=\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2}{\mathrm{I} \omega^{2}}}=\frac{3}{4} \end{aligned}

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