Question

A particle of mass $m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $a$. A force $\overrightarrow{F}=y\hat{i}-x\hat{j}$ newton acts on the particle, where $x,y$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A ($a,0$) to point B ($0,a$) along the circular path is

• A. $\frac{\pi a^2}{2}J$
• B. $\frac{\pi a^2}{4}J$
• C. $Noneofthese$
• D. $\frac{\pi a^2}{3}J$

Answer 1

Answer: (D)

$W=\int F.dr=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})$

$=\int (ydx-xdy)$ .........(1)

$\because x^2+y^2=a^2\therefore xdx+ydy=0$

$\Rightarrow W=\int (y\left(\frac{-ydy}{x}\right)-xdy)=-\int \frac{x^2+y^2}{x}dy$

$=-{\int }_0^a\frac{a^2}{\sqrt{a^2-y^2}}dy=-\frac{\pi a^2}{2}$