A farmer moves along the boundary of a square field of side $10\mathit{ m}$ in $40\mathit{ s}$, the displacement of the farmer at the end of $2\mathit{ min }20\mathit{ s}$ is __________$m$.

The farmer covers $40m$ in $40s$, then in $2\mathit{ min }20\mathit{ s}$ he will cover,
$2\times 60+20=120+20$
                      $=140\mathit{ s}$
In $140\mathit{ s}$ he will cover,
$\frac{140}{40}=3.5$
In $140\mathit{ s}$ The farmer will cover three and a half rounds of the field.
So, at the end of $140\mathit{ s}$ farmer will be diagonally opposite to where he started. If ABCD is a square, then using Pythagoras theorem
Displacement of the farmer is equal to,
     $\mathit{AC}=\sqrt{{\mathit{AB}}^2+{\mathit{BC}}^2}$
$\Rightarrow \mathit{AC}=\sqrt{{10}^2+{10}^2}$
$\Rightarrow \mathit{AC}=10\sqrt{2}\mathit{ m}$
$\Rightarrow \mathit{AC}=14.14\mathit{ m}$
The farmer’s displacement after $2\mathit{ min }20\mathit{ s}$ will be $14.14\mathit{ m}$.