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The work done in rotating a dipole placed parallel to the electric field through $180^{\circ}$ is 'w'. So the work done in rotating it through $60^{\circ}$ is $\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1\right)$
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The correct answer is:
$\frac{\mathrm{w}}{4}$
Work done $\mathrm{W}=\mathrm{p} E\left[\cos \theta_0-\cos \theta\right]$ Given, $\mathrm{W}=\mathrm{pE}\left(1-\cos 180^{\circ}\right)$
$\therefore \quad \mathrm{W}=\mathrm{pE}\left(1-\cos 180^{\circ}\right)$
$\begin{aligned} & \mathrm{W}=\mathrm{pE}(1-(-1)) \\ & \mathrm{W}=2 \mathrm{pE} \\ & \mathrm{pE}=\frac{1}{2} \mathrm{~W}\end{aligned}$
$\therefore \quad$ When $\theta=60^{\circ}$
$\begin{aligned} & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\cos 60^{\circ}\right) \\ & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\frac{1}{2}\right) \\ & \mathrm{W}^{\prime}=\frac{1}{2} \mathrm{pE} \\ & \mathrm{W}^{\prime}=\frac{1}{2} \times \frac{1}{2} \mathrm{~W} \\ & \mathrm{~W}^{\prime}=\frac{1}{4} \mathrm{~W}\end{aligned}$
$\therefore \quad \mathrm{W}=\mathrm{pE}\left(1-\cos 180^{\circ}\right)$
$\begin{aligned} & \mathrm{W}=\mathrm{pE}(1-(-1)) \\ & \mathrm{W}=2 \mathrm{pE} \\ & \mathrm{pE}=\frac{1}{2} \mathrm{~W}\end{aligned}$
$\therefore \quad$ When $\theta=60^{\circ}$
$\begin{aligned} & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\cos 60^{\circ}\right) \\ & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\frac{1}{2}\right) \\ & \mathrm{W}^{\prime}=\frac{1}{2} \mathrm{pE} \\ & \mathrm{W}^{\prime}=\frac{1}{2} \times \frac{1}{2} \mathrm{~W} \\ & \mathrm{~W}^{\prime}=\frac{1}{4} \mathrm{~W}\end{aligned}$
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