A bar of uniform cross section and weighing 100 N is held horizontally using two massless and inextensible strings S1 and S2 as shown in the figure.

The tensions in the strings are

The correct option is: T₁ = 0 N and T₂ = 100 N.

1. Analysis of the Diagram:
By inspecting the provided image, we can identify the following parameters and layout:
- A horizontal Bar of uniform cross section is supported by two vertical strings, S1 and S2, suspended from a Rigid support.
- String S1 with tension T₁ is attached at the leftmost end of the bar.
- String S2 with tension T₂ is attached at the center of the bar. The diagram shows two equal length segments of L/2 at the bottom, which confirms that string S2 is located exactly at the midpoint of the bar of total length $L$.
- The bar is uniform, meaning its center of gravity lies exactly at its geometric center (midpoint). Therefore, the weight of the bar ($W=100\text{N}$) acts vertically downwards at the midpoint.

2. Rotational Equilibrium:
For the bar to remain horizontal and in static equilibrium, the sum of all torques about any pivot point must be zero. Let us compute the torque about the midpoint of the bar:

$\sum \tau =0$

Taking the midpoint as the pivot:
- The downward gravitational force (weight $W$) acts at the midpoint, so its torque is zero.
- The tension $T_2$ is applied at the midpoint, so its torque is zero.
- The tension $T_1$ is applied at the leftmost end, at a distance of $\frac{L}{2}$ from the pivot.

Thus, the torque equation about the midpoint yields:

$T_1\times \frac{L}{2}=0$

Since the length $L$ of the bar is non-zero, it follows directly that:

$T_1=0\text{N}$

3. Translational Equilibrium:
For the bar to be in vertical translational equilibrium, the net vertical force must be zero. The sum of upward forces must equal the downward force (weight of the bar):

$T_1+T_2=W$

Substituting $T_1=0\text{N}$ and $W=100\text{N}$ into the equation:

$0+T_2=100\text{N}$

$T_2=100\text{N}$

Conclusion:
The tensions in the strings are T₁ = 0 N and T₂ = 100 N.