If 10 J of work is done to turn a vertically suspended rod through 60⁰ from its mean position. The additional work to be done to turn it further through 120⁰ is

Work done to turn a vertically suspended rod through an angle $\theta$ is equal to mgl (1-cos$\theta$)

$$\begin{gathered} 10=mgl(1-\cos {60}^0)=\frac{mgl}{2} \\ mgl=20 \end{gathered}$$

here additional work done from  ${60}^0 to {120}^0$is

$$\begin{gathered} mgl(1-\cos {120}^0-1+\cos {60}^0) \\ mgl(0.5+0.5)=mgl=20J \end{gathered}$$