A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity ω about a vertical axis passing through one end. The tension in the rod at a distance x from the axis is

The correct option is:
$\frac{1}{2}m{\omega }^2\frac{(l^2-x^2)}{l}$

Step-by-Step Derivation:

1. Understand the system:
We have a uniform rod of mass m and length l rotating in a horizontal plane with a constant angular velocity ω about a vertical axis passing through one of its ends (let this end be at position y = 0). We want to find the tension T in the rod at a distance x from this axis of rotation.

2. Mass per unit length:
Since the rod is uniform, its linear mass density (mass per unit length), denoted by λ, is constant:
$\lambda =\frac{m}{l}$

3. Consider a small element of the rod:
Let us consider a small element of length dy at a distance y from the axis of rotation, where y is between x and l.
The mass of this small element, dm, is given by:
$dm=\lambda dy=\frac{m}{l}dy$

4. Centripetal force on the small element:
This element is moving in a circular path of radius y with angular velocity ω. The centripetal force required to keep this element in circular motion is:
$dF=dm\cdot {\omega }^2\cdot y=\left(\frac{m}{l}dy\right){\omega }^{2}y$

5. Relating centripetal force to tension:
The tension in the rod is what provides the necessary centripetal force. Let the tension at distance y be T and at distance y + dy be T + dT.
The net inward radial force on the element dy is:
$T-(T+dT)=dF$
$-dT=\frac{m}{l}{\omega }^{2}ydy$

6. Integrating to find the total tension:
To find the tension T at a distance x, we integrate from y = x (where the tension is T) to the free end of the rod at y = l (where the tension is 0 because there is no mass beyond the outer end to pull on it):
${\int }_T^0-dT={\int }_x^l\frac{m}{l}{\omega }^{2}ydy$
$-{\left[T\right]}_T^0=\frac{m{\omega }^2}{l}{\left[\frac{y^2}{2}\right]}_x^l$
$T=\frac{m{\omega }^2}{2l}(l^2-x^2)$

This can be rewritten in the standard form matching the correct option:
$T=\frac{1}{2}m{\omega }^2\frac{(l^2-x^2)}{l}$