JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Work, Power & Energy

The work done on a particle of mass

m

by a force,

K [\frac{x}{( x ^{2} + y ^{2} ) ^{3/2}} \hat{i} + \frac{y}{( x ^{2} + y ^{2} ) ^{3/2}} \hat{j}]

(

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) $\frac{2 K π}{a}$
(B) $\frac{K π}{a}$
(C) $\frac{K π}{2 a}$
(D)0✓ Correct

Answer: (D)

Solution

We check if the force is conservative by verifying

\frac{\partial F _{x}}{\partial y} = \frac{\partial F _{y}}{\partial x}

.

Since this condition is satisfied, the force is conservative.

For a conservative force, the work done depends only on the initial and final positions, not on the path.

On the circular path

x^{2} + y^{2} = a^{2}

, we have:

W = \int_{a}^{0} F_{x} d x + \int_{0}^{a} F_{y} d y = \frac{K}{a ^{3}} \int_{a}^{0} x d x + \frac{K}{a ^{3}} \int_{0}^{a} y d y

= \frac{K}{a ^{3}} [\frac{x ^{2}}{2}]_{a}^{0} + \frac{K}{a ^{3}} [\frac{y ^{2}}{2}]_{0}^{a}

= \frac{K}{a ^{3}} (- \frac{a ^{2}}{2} + \frac{a ^{2}}{2}) = 0

Hence the answer is (D).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Work, Power & Energy

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