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Work Power and Energy question

2018 · Shift 0 · Q61
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Work Power and Energy question

2018 · Shift 0 · Q61

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A particle is moving in a circular path of radius aaa under the action of an attractive potential U=−k2r2U = - {k \over {2{r^2}}}U=−2r2k​ Its total energy is:
  1. A
    −32ka2- {3 \over 2}{k \over {{a^2}}}−23​a2k​
  2. B
    Zero
  3. C
    −k4a2- {k \over {4{a^2}}}−4a2k​
  4. D
    k2a2{k \over {2{a^2}}}2a2k​
View written solutionFree

Correct answer: B

  1. Given potential

The attractive central potential is

U(r)=−k2r2.U(r)=-\frac{k}{2r^2}.U(r)=−2r2k​.

For a circular orbit of radius aaa, we use the condition that the required centripetal force is provided by the central force.

  1. Find the force from the potential

The radial force is

F(r)=−dUdr.F(r)=-\frac{dU}{dr}.F(r)=−drdU​.

Now,

U(r)=−k2r−2U(r)=-\frac{k}{2}r^{-2}U(r)=−2k​r−2

so

dUdr=−k2(−2)r−3=kr3.\frac{dU}{dr}=-\frac{k}{2}(-2)r^{-3}=\frac{k}{r^3}.drdU​=−2k​(−2)r−3=r3k​.

Hence,

F(r)=−kr3.F(r)=-\frac{k}{r^3}.F(r)=−r3k​.

This is attractive, as expected.

At r=ar=ar=a,

F(a)=−ka3.F(a)=-\frac{k}{a^3}.F(a)=−a3k​.

Its magnitude is

ka3.\frac{k}{a^3}.a3k​.
  1. Apply circular motion condition

For circular motion,

mv2a=ka3.\frac{mv^2}{a}=\frac{k}{a^3}.amv2​=a3k​.

Therefore,

mv2=ka2mv^2=\frac{k}{a^2}mv2=a2k​

and so the kinetic energy is

T=12mv2=k2a2.T=\frac12 mv^2=\frac{k}{2a^2}.T=21​mv2=2a2k​.
  1. Potential energy at radius aaa
U(a)=−k2a2.U(a)=-\frac{k}{2a^2}.U(a)=−2a2k​.
  1. Total energy
E=T+U=k2a2−k2a2=0.E=T+U=\frac{k}{2a^2}-\frac{k}{2a^2}=0.E=T+U=2a2k​−2a2k​=0.
  1. Check options
  • A: −32ka2-\dfrac{3}{2}\dfrac{k}{a^2}−23​a2k​ — incorrect
  • B: 000 — correct
  • C: −k4a2-\dfrac{k}{4a^2}−4a2k​ — incorrect
  • D: k2a2\dfrac{k}{2a^2}2a2k​ — incorrect

Therefore, the correct option is B.

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