A rod of length L is pivoted at one of its ends and rotated in a...

Let the mass per unit length of the rod be $\lambda$ and its angular velocity be $\omega$.

Consider an element of length $dx$ at a distance $x$ from the pivoted end. The centripetal force required for this element is $dF = (\lambda dx) \omega^2 x$.

The tension $T(x)$ at a distance $x$ from the pivot provides the necessary centripetal force for the remaining portion of the rod from $x$ to $L$.

$T(x) = \int_x^{L} \lambda \omega^2 x' dx' = \dfrac{\lambda \omega^2}{2} (L^2 - x^2)$

Since $T(x)$ is a decreasing function of $x$, the tension is greater at points closer to the pivot.

Given $x_1 = \dfrac{L}{4}$ and $x_2 = \dfrac{3L}{4}$, we have $x_1 < x_2$.

Therefore, $T_1 > T_2$.