Why is the handle of a screwdriver made long to rotate a screw?

$A$ solid cone hangs from a frictionless pivot at the origin $O$,as shown. If $\hat{i}$,$\hat{j}$,and $\hat{k}$ are unit vectors,and $a, b$,and $c$ are positive constants,which of the following forces $\vec{F}$ applied to the rim of the cone at a point $P$ results in a torque $\vec{\tau}$ on the cone with a negative $z$-component $\tau_z$?

$A$ 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.

$A$ force $\overrightarrow{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ acts at a point $\overrightarrow{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point $\overrightarrow{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is:

The moment of the force,$\overrightarrow{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at point $(2, 0, -3)$,about the point $(2, -2, -2)$,is given by

$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)