One end of a uniform rod of mass m₁ and cross-sectional area A is hung from a ceiling. The other end of the bar is supporting mass m₂. The stress at the midpoint is

The correct option is: g(2m₂ + m₁)/2A (which can be rewritten as $\frac{g(2m_2+m_1)}{2A}$).

Let us derive the tension and stress at the midpoint of the rod step-by-step.

Step 1: Understand the system and forces
We are given:
- A uniform rod of mass $m_1$, length $L$, and cross-sectional area $A$ suspended vertically from a ceiling.
- A block of mass $m_2$ is supported at the lower end of the rod.
- We need to find the stress at the midpoint of the rod.

Step 2: Determine the mass supported by the midpoint
The midpoint of the rod divides it into two equal halves of length $\frac{L}{2}$ each.
Since the rod is uniform, the mass of the lower half of the rod is:
$m_{\text{lower}}=\frac{m_1}{2}$

The total mass hanging below the midpoint consists of:
1. The mass of the lower half of the rod itself, which is $\frac{m_1}{2}$.
2. The load mass $m_2$ attached to the bottom end of the rod.

Therefore, the total downward force (tension $T$) acting at the midpoint of the rod is the weight of all the mass below it:
$T=(m_2+\frac{m_1}{2})g$
We can simplify this expression by taking a common denominator of 2:
$T=\frac{(2m_2+m_1)g}{2}$

Step 3: Calculate the stress at the midpoint
Stress is defined as the restoring force (or tension) per unit cross-sectional area:
$\text{Stress}=\frac{T}{A}$

Substituting the expression for tension $T$ at the midpoint into the stress formula:
$\text{Stress}=\frac{\frac{(2m_2+m_1)g}{2}}{A}$
$\text{Stress}=\frac{g(2m_2+m_1)}{2A}$

This matches the correct option.