A uniform rope is supported by two level pin support as shown in the figure. Mass of the rope is m. Find the tension at mid-point.

The correct option is (mg√3)/2.

Let us analyze the forces acting on the uniform rope of mass $m$.

From the provided image, we can observe that the rope is supported at both ends by two level pin supports, and it makes an angle of $30°$ with the horizontal dotted line at both supports.

Let $T_0$ be the tension in the rope at the supports. We can resolve this tension into vertical and horizontal components:

1. The vertical component of tension at each support is:
$T_y=T_{0}sin(30°)$
2. The horizontal component of tension at each support is:
$T_x=T_{0}cos(30°)$

For the vertical equilibrium of the entire rope of mass $m$, the total upward vertical force exerted by both supports must equal the total downward gravitational force ($mg$):
$2T_{0}sin(30°)=mg$
Substituting the value $sin(30°)=\frac{1}{2}$:
$2T_0\left(\frac{1}{2}\right)=mg$
$T_0=mg$

Now, let us find the tension at the mid-point (the lowest point of the rope). At the mid-point, the tangent to the rope is perfectly horizontal. Therefore, the tension at this point, $T_{\text{mid}}$, is entirely horizontal.

Since there are no external horizontal forces acting on the rope, the horizontal component of the tension must be constant at every point along the rope:
$T_{\text{mid}}=T_x=T_{0}cos(30°)$

Substituting the values of $T_0=mg$ and $cos(30°)=\frac{\sqrt{3}}{2}$:
$T_{\text{mid}}=mg\left(\frac{\sqrt{3}}{2}\right)=\frac{mg\sqrt{3}}{2}$

Therefore, the tension at the mid-point of the rope is $\frac{mg\sqrt{3}}{2}$.