A vertical hanging bar of length l and mass m per unit length carries a load of mass M at the lower end, its upper end is clamped to a rigid support. The tensile force at a distance x from support is

The correct option is Mg + mg(l - x).

To find the tensile force at a distance $x$ from the rigid support, we need to analyze the forces acting on the section of the bar that lies below this point.

1. Identify the length of the lower section:
The total length of the hanging bar is $l$. The point under consideration is at a distance $x$ from the top support. Therefore, the length of the bar suspended below this point is:
$l_{\text{lower}}=l-x$

2. Determine the mass of this lower section:
The mass per unit length of the bar is $m$. Thus, the mass of the section of length $(l-x)$ is:
$m_{\text{lower}}=m(l-x)$

3. Calculate the total suspended mass:
A load of mass $M$ is attached at the lower end of the bar. The total mass suspended below the point at distance $x$ is the sum of the mass of the lower portion of the bar and the load $M$:
$M_{\text{total}}=M+m(l-x)$

4. Find the tensile force:
The tensile force (tension) $T$ at a distance $x$ from the support supports the weight of the entire mass suspended below it in static equilibrium:
$T=M_{\text{total}}g$
Substituting the value of $M_{\text{total}}$ into the equation:
$T=[M+m(l-x\left)\right]g$
$T=Mg+mg(l-x)$