JEE Advanced · Physics · 6. Work Power Energy
The work done on a particle of mass m by a force (K being a constant of appropriate dimensions, when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is
- A
- B
- C
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
\(d w==\vec{F} \cdot d \vec{r}=\vec{F} \cdot(d x \hat{i}+d y \hat{j})=K \int\) \(\frac{x d x}{\left(x^2+y^2\right)^{3 / 2}}+\frac{y d y}{\left(x^2+y^2\right)^{3 / 2}}\)
\(x^2+y^2=a^2\)
\(w =\frac{ K }{ a ^3} \int_{ a }^0 xdx +\int_0^{ a } ydy =\frac{ K }{ a ^3}\left(\frac{- a ^2}{2}+\frac{ a ^2}{2}\right)=0\).
\(x^2+y^2=a^2\)
\(w =\frac{ K }{ a ^3} \int_{ a }^0 xdx +\int_0^{ a } ydy =\frac{ K }{ a ^3}\left(\frac{- a ^2}{2}+\frac{ a ^2}{2}\right)=0\).
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