A and B are two wires. The radius of A is twice that of B. they are stretched by the same load. Then the stress on B is

The correct option is Four times that on A.

To find the relationship between the stresses in the two wires, we can break down the physics principles involved step-by-step.

Step 1: Formula for Stress
Stress is defined as the force applied per unit cross-sectional area of a wire:
$\mathrm{Stress}=\frac{\mathrm{Force}}{\mathrm{Area}}$
For a cylindrical wire with a circular cross-section of radius $r$, the area is:
$\mathrm{Area}=\pi r^2$
Therefore, the stress $\sigma$ can be written as:
$\sigma =\frac{F}{\pi r^2}$

Step 2: Set up the Given Conditions
Let $r_B$ be the radius of wire B. The radius of wire A is twice that of B, so:
$r_A=2r_B$
Both wires are stretched by the same load, which means the stretching force $F$ is identical for both wires:
$F_A=F_B=F$

Step 3: Compare the Stresses
Let us calculate the stress on wire A (${\sigma }_A$) and wire B (${\sigma }_B$):
For wire A:
${\sigma }_A=\frac{F}{\pi r_A^2}=\frac{F}{\pi {\left(2r_B\right)}^2}=\frac{F}{4\pi r_B^2}$
For wire B:
${\sigma }_B=\frac{F}{\pi r_B^2}$
Now, comparing ${\sigma }_B$ with ${\sigma }_A$:
${\sigma }_B=4\times \left(\frac{F}{4\pi r_B^2}\right)=4{\sigma }_A$

Thus, the stress on wire B is four times that on wire A because the cross-sectional area of wire B is four times smaller than that of wire A.