1. Set up the rod element

Consider a uniform rod of length $\ell$, rotating with constant angular speed $\omega$ about one end in a horizontal plane.

Let:

• linear mass density $= \lambda = \dfrac{M}{\ell}$
• $T(x)$ = internal tension in the rod at a point at distance $x$ from the axis

We need how $T(x)$ varies with $x$.

2. Interpretation of tension at point $x$

The tension at section $x$ must provide the centripetal force needed for the part of the rod beyond that section, i.e. from $x$ to $\ell$.

Take a small element of length $dr$ at distance $r$ from the axis.

Its mass is $dm = \lambda \, dr$

Required centripetal force is $dF = dm\, \omega^2 r = \lambda \omega^2 r\, dr$

So the tension at $x$ is $T(x) = \int_x^{\ell} \lambda \omega^2 r\, dr$

3. Integrate

$T(x) = \lambda \omega^2 \int_x^{\ell} r\, dr$

$T(x) = \lambda \omega^2 \left[\frac{r^2}{2}\right]_x^{\ell}$

$T(x) = \frac{\lambda \omega^2}{2}(\ell^2 - x^2)$

4. Nature of the graph

Thus, $T(x) \propto \ell^2 - x^2$

This means:

• maximum at $x=0$: $T(0)=\frac{\lambda \omega^2 \ell^2}{2}$
• zero at $x=\ell$: $T(\ell)=0$
• decreases with $x$
• variation is parabolic, concave downward

Also, $\frac{dT}{dx} = -\lambda \omega^2 x$

So slope is zero at $x=0$ and becomes more negative as $x$ increases.

Hence the graph starts from a maximum with horizontal tangent at $x=0$, then falls parabolically to zero at $x=\ell$.

5. Match with the options

The correct graph is the one representing $T(x)=\frac{\lambda \omega^2}{2}(\ell^2-x^2)$ which is a downward-opening parabola from $(0, T_{\max})$ to $(\ell,0)$.

Therefore, the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So they agree.