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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

The work done on a particle of mass m by a force, K[x/(x²+y²)^(3/2) î + y/(x²+y²)^(3/2) ĵ] (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 [JEE Advance 2013]

Answer: (D)

Solution

Writing the position vector r = xî + yĵ with |r| = a (constant on the circular path), the force simplifies to F = K(xî+yĵ)/(x²+y²)^(3/2) = K·r/|r|³ = (K/a²)·r̂, which is purely radial (central). Since the

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Question ID VK-154215