$A$ door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give a reason for your answer.

(N/A) No,the weight of the door does not cause any torque about the vertical axis of rotation.
The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
In this case,the vertical axis of rotation is the $Y$-axis. The weight $W$ of the door acts vertically downwards,which is along the negative $Y$-axis (i.e.,$\vec{F} = -W \hat{j}$).
The position vector $\vec{r}$ of any point on the door lies in the $XY$ plane. Since the force vector $\vec{F}$ is parallel to the axis of rotation ($Y$-axis),the line of action of the weight passes through the axis of rotation.
Mathematically,the perpendicular distance from the axis of rotation to the line of action of the force is zero. Therefore,the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$ about the vertical axis.