As shown in figure, a stick half of which is made of wood and other half of iron is pivoted as shown in fig. (a) It is pivoted at wooden end and force is applied to the other end of it at right angles to its length. (b) It is pivoted at iron end and the same force is applied to the other end at right angles to its length. Then angular acceleration produced is

Wood is less dense than steel.
We know that 

$\mathrm{\tau } = \mathrm{I} \times \mathrm{\alpha }$

I here stand for moment of inertia.

τ  is torque, α is the angular acceleration, and

$\mathrm{\tau }=\mathrm{I\alpha }=\mathrm{constant}$

$\Rightarrow \mathrm{\alpha }=\frac{\mathrm{\tau }}{\mathrm{I}}$

$\mathrm{\alpha }\propto \frac{1}{\mathrm{I}}$

Since ${\mathrm{I}}_1>{\mathrm{I}}_2$ and is the same in both circumstances, the result is ${\mathrm{\alpha }}_1<{\mathrm{\alpha }}_2$.

When I is small, angular acceleration will be greater; hence, lighter materials (such as wood) should be placed farther from the axis of rotation, especially when the stick is pivoted at the steel end.

Hence the correct answer is More in case (b) than in (a).