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(D) Consider a small element of length $dx$ at a distance $r$ from the axis of rotation. The mass of this element is $dm = (M/l) dr$,where $M$ is the

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(D) Consider a small element of length $dx$ at a distance $r$ from the axis of rotation. The mass of this element is $dm = (M/l) dr$,where $M$ is the total mass of the rod.The tension $T(x)$ at a distance $x$ from the axis must provide the centripetal force required to rotate the portion of the rod from $x$ to $l$.$T(x) = \int_{x}^{l} (dm) \omega^2 r = \int_{x}^{l} \left(\frac{M}{l}\right) dr \omega^2 r$$T(x) = \frac{M \omega^2}{l} \int_{x}^{l} r dr = \frac{M \omega^2}{l} \left[ \frac{r^2}{2} \right]_{x}^{l}$$T(x) = \frac{M \omega^2}{2l} (l^2 - x^2)$This equation represents a downward-opening parabola. At $x = 0$,$T(0) = \frac{1}{2} M \omega^2 l$. At $x = l$,$T(l) = 0$. The graph that matches this parabolic decay is shown in option $D$.

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