Question

Suppose that a rigid body is free to rotate about a fixed point $O$. If a force $\overrightarrow{F}$ acts on the body at a point $P$, then the body will tend to rotate about an axis through $O$. This effect is measured by the torque vector $t$ which is given by $\overrightarrow{t}=\overrightarrow{r}\times \overrightarrow{F}$. Where $\overrightarrow{r}$ is the position vector $\overrightarrow{OP}$. The straight line through $O$ determined by $\overrightarrow{t}$ is the axis of rotation. The vector $\overrightarrow{r},\overrightarrow{F}$ and $\overrightarrow{T}$ form a right handed system.
The magnitude of the torque $\left(\text{in units}\right)$ exerted at origin by the force $\overrightarrow{F}=\hat{i}+2\hat{j}+\hat{k}$ at the point $(1,1,1)$ is

• A. $\sqrt{7}$
• B. $\sqrt{5}$
• C. $\sqrt{3}$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

Given: $\overrightarrow{F}=\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{r}=\overrightarrow{OP}=(1-0)\hat{i}+(1-0)\hat{j}+(1-0)\hat{k}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{t}=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{t}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 1\end{array}\right|=(1-2)\hat{i}-(1-1)\hat{j}+(2-1)\hat{k}$
$\overrightarrow{t}=-\hat{i}+\hat{k}$
Magnitude of $\overrightarrow{t}$
$|\overrightarrow{t}|=\sqrt{1^2+1^2}=\sqrt{2}\text{units}$