In the figure (a),half of the meter scale is | vedclass

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(B) The angular acceleration $\alpha$ is given by the relation $	au = I \alpha$,where $	au$ is the torque and $I$ is the moment of inertia. Thus,$\a

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more angular acceleration will be produced in $(b)$

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(B) The angular acceleration $\alpha$ is given by the relation $	au = I \alpha$,where $	au$ is the torque and $I$ is the moment of inertia. Thus,$\alpha = \frac{	au}{I}$.In both cases,the torque $	au = F 	imes L$ is the same,where $L$ is the total length of the scale.Let $m_w$ be the mass of the wooden half and $m_s$ be the mass of the steel half. Since steel is denser than wood,$m_s > m_w$.In case $(a)$,the pivot is at the center of the wooden part. The moment of inertia $I_a$ is the sum of the moment of inertia of the wooden part about $O$ and the steel part about $O$.In case $(b)$,the pivot is at the center of the steel part. The moment of inertia $I_b$ is the sum of the moment of inertia of the steel part about $O'$ and the wooden part about $O'$.Since the heavier mass (steel) is closer to the pivot in case $(b)$,the moment of inertia $I_b$ is less than $I_a$ $(I_b Since $\alpha \propto \frac{1}{I}$,a smaller moment of inertia results in a larger angular acceleration. Therefore,more angular acceleration will be produced in $(b)$.

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