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(B) According to the law of conservation of energy,the loss in potential energy of the weight is equal to the gain in the total kinetic energy of the

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$\frac{2 mgh}{I + mr^2}$

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(B) According to the law of conservation of energy,the loss in potential energy of the weight is equal to the gain in the total kinetic energy of the system (rotational kinetic energy of the wheel + translational kinetic energy of the weight).Loss in potential energy = $mgh$Gain in kinetic energy = $\frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$Since the cord does not slip,the linear velocity $v$ of the weight is related to the angular velocity $\omega$ of the wheel by $v = \omega r$.Substituting $v = \omega r$ into the energy equation:$mgh = \frac{1}{2} I \omega^2 + \frac{1}{2} m(\omega r)^2$$mgh = \frac{1}{2} (I + mr^2) \omega^2$Solving for $\omega^2$:$\omega^2 = \frac{2 mgh}{I + mr^2}$

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