Question

A force $\overrightarrow{F}=b\frac{-yi+xj}{x^2+y^2}\text{N}$ ($b$ is a constant) acts on a particle as it undergoes counterclockwise circular motion in a circle $x^2+y^2=16$.
The work done by the force when the particle undergoes one complete revolution is ($x,y$ are in $m$)

• A. Zero
• B. $2\pi b\text{J}$
• C. $2b\text{J}$
• D. None of these

Answer 1

The correct option is B $2\pi b\text{J}$

For complete revolution,
work done can be calculated as,
$W=\oint \overrightarrow{F}.\overrightarrow{dr}$
$W=\oint \overrightarrow{F}.(dx\hat{i}+dy\hat{j})$
$W=\oint b\frac{-yi+xj}{x^2+y^2}.(dx\hat{i}+dy\hat{j})$
$W=\oint b\frac{-ydx+xdy}{x^2+y^2}$
Given path is $x^2+y^2=16,$ here we can consider paramteric points as $x=4\cos \theta ,y=4\sin \theta \Rightarrow dx=-4\sin \theta d\theta ,dy=4\cos \theta d\theta$

$W=\oint b\frac{-4\sin \theta dx+4\cos \theta dy}{16}$
$W=\oint b\frac{16{\sin }^2\theta d\theta +16{\cos }^2\theta d\theta }{16}$
$W=\oint bd\theta =b\left(2\pi \right)=2b\pi$