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Work Power and Energy question

2013 · Shift 1 · Q43
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Work Power and Energy question

2013 · Shift 1 · Q43

JEE AdvancedPhysicsWork Power and EnergyMCQ+2 / −0.5
The work done on a particle of mass m by a force K[x(x2+y2)3/2i^+y(x2+y2)3/2j^]K\left[ {{x \over {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}\widehat i + {y \over {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}\widehat j} \right]K[(x2+y2)3/2x​i+(x2+y2)3/2y​j​](K being a constant of appropriate dimensions), when the particle is taken from the point (a,0)\left( {a,0} \right)(a,0) to the point (0,a)\left( {0,a} \right)(0,a) along a circular path of radius a about the origin in the x-y plane is
  1. A
    2Kπa{{2K\pi } \over a}a2Kπ​
  2. B
    Kπa{{K\pi } \over a}aKπ​
  3. C
    Kπ2a{{K\pi } \over {2a}}2aKπ​
  4. D
    0
View written solutionFree

Correct answer: D

  1. Given force field

The force is

F⃗=K[x(x2+y2)3/2i^+y(x2+y2)3/2j^].\vec F=K\left[\frac{x}{(x^2+y^2)^{3/2}}\hat i+\frac{y}{(x^2+y^2)^{3/2}}\hat j\right].F=K[(x2+y2)3/2x​i^+(x2+y2)3/2y​j^​].

We need the work done from (a,0)(a,0)(a,0) to (0,a)(0,a)(0,a) along the circular arc of radius aaa centered at the origin.

  1. Identify the path

For a circle of radius aaa about the origin,

x2+y2=a2.x^2+y^2=a^2.x2+y2=a2.

So along the path,

(x2+y2)3/2=(a2)3/2=a3.(x^2+y^2)^{3/2}=(a^2)^{3/2}=a^3.(x2+y2)3/2=(a2)3/2=a3.

Hence on the path,

F⃗=K(xa3i^+ya3j^)=Ka3(xi^+yj^).\vec F=K\left(\frac{x}{a^3}\hat i+\frac{y}{a^3}\hat j\right) =\frac{K}{a^3}(x\hat i+y\hat j).F=K(a3x​i^+a3y​j^​)=a3K​(xi^+yj^​).

But (xi^+yj^)(x\hat i+y\hat j)(xi^+yj^​) is the position vector, which on the circle points radially outward.

  1. Direction of displacement along the circular arc

The displacement element along the circular path is tangential:

dr⃗=dx i^+dy j^.d\vec r = dx\,\hat i + dy\,\hat j.dr=dxi^+dyj^​.

For motion on a circle, the tangent is perpendicular to the radius vector. Therefore,

(xi^+yj^)⋅dr⃗=x dx+y dy.(x\hat i+y\hat j)\cdot d\vec r = x\,dx+y\,dy.(xi^+yj^​)⋅dr=xdx+ydy.

But since

x2+y2=a2,x^2+y^2=a^2,x2+y2=a2,

differentiating gives

2x dx+2y dy=0⇒x dx+y dy=0.2x\,dx+2y\,dy=0 \quad\Rightarrow\quad x\,dx+y\,dy=0.2xdx+2ydy=0⇒xdx+ydy=0.

Thus,

F⃗⋅dr⃗=Ka3(x dx+y dy)=0.\vec F\cdot d\vec r=\frac{K}{a^3}(x\,dx+y\,dy)=0.F⋅dr=a3K​(xdx+ydy)=0.
  1. Compute the work

Work done is

W=∫F⃗⋅dr⃗.W=\int \vec F\cdot d\vec r.W=∫F⋅dr.

Since the integrand is zero everywhere on the circular arc,

W=0.W=0.W=0.
  1. Check options
  • A: 2Kπa\dfrac{2K\pi}{a}a2Kπ​ ❌
  • B: Kπa\dfrac{K\pi}{a}aKπ​ ❌
  • C: Kπ2a\dfrac{K\pi}{2a}2aKπ​ ❌
  • D: 000 ✅

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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