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MHT CET 2023 (13 May, Shift 2) · Electric Dipole

Electric Dipole question from MHT CET 2023 (13 May, Shift 2)

Q46MHT CET 2023 · 13 MAY · SHIFT 2Medium

Electric Dipole

The work done in rotating a dipole placed parallel to the electric field through 180∘180^\circ is W. So, the work done in rotating it through 60∘60^\circ is (cos⁡0∘=1, cos⁡60∘=12, cos⁡180∘=−1)\left(\cos 0^\circ = 1,\ \cos 60^\circ = \frac{1}{2},\ \cos 180^\circ = -1\right)

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Answer: (d)

  1. Wθ=pE(1−cos⁡θ)W_\theta = pE(1 - \cos\theta) from the parallel position.
  2. 180∘180^\circ: W=2pEW = 2pE; 60∘60^\circ: W′=pE2W' = \frac{pE}{2}
  3. W′=W4W' = \frac{W}{4}
  4. So the answer is (d) W/4W/4.
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