A pendulum of length $L$ oscillates with an angular amplitude of $\theta = 60^\circ$ and time period $T$. Let $T_0 = 2\pi\sqrt{\dfrac{L}{g}}$ be the time period for small angle of oscillations, where $g$ is the acceleration due to gravity. If air resistance is negligibly small and the string remains straight, then which one of the following is correct?

The restoring torque acting on the pendulum is $\tau = -mgL \sin\theta$.

The angular acceleration is $\alpha = -\dfrac{g}{L} \sin\theta$.

For small angular amplitudes, $\sin\theta \approx \theta$, which gives the time period $T_0 = 2\pi\sqrt{\dfrac{L}{g}}$.

For a large angular amplitude ($\theta = 60^\circ$), $\sin\theta < \theta$. This implies that the restoring torque is smaller in magnitude compared to the linear approximation $\tau = -mgL\theta$.

Due to the smaller restoring torque, the angular acceleration is reduced, which increases the time taken to complete an oscillation.

The exact time period for large amplitudes is given by the series expansion $T = T_0 \left( 1 + \dfrac{1}{4}\sin^2\left(\dfrac{\theta_0}{2}\right) + \dots \right)$.

Since $\theta_0 = 60^\circ$, the terms in the bracket are positive, yielding $T > T_0$.

Therefore, $T$ will be slightly greater than $T_0$.

Answer: $T$ will be slightly greater than $T_0$.