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 \(\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1\right)\)

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 \(\theta=60^{\circ}\)
\(\begin{aligned} & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\cos 60^{\circ}\right) \\ & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\frac{1}{2}\right) \\ & \mathrm{W}^{\prime}=\frac{1}{2} \mathrm{pE} \\ & \mathrm{W}^{\prime}=\frac{1}{2} \times \frac{1}{2} \mathrm{~W} \\ & \mathrm{~W}^{\prime}=\frac{1}{4} \mathrm{~W}\end{aligned}\)