The wires A and B shown in figure are made of the same material and have radii r_A and r_B, respectively. The block between them has a mass m. When the force F is mg/3, one of the wires breaks. Then

Tension in $\mathrm{B}={\mathrm{T}}_{\mathrm{B}}=\frac{\mathrm{mg}}{3}$

Tension in $\mathrm{A}={\mathrm{T}}_{\mathrm{A}}={\mathrm{T}}_{\mathrm{B}}+\mathrm{mg}=\frac{4\mathrm{mg}}{3}$

$\therefore {\mathrm{T}}_{\mathrm{A}}=4{\mathrm{T}}_{\mathrm{B}}$

Stress $\frac{\mathrm{T}}{{\mathrm{\pi r}}^2}=\mathrm{S}$

Wire breaks when S is equal to breaking stress

For ${\mathrm{r}}_{\mathrm{A}}={\mathrm{r}}_{\mathrm{B}},{\mathrm{S}}_{\mathrm{A}}=4{\mathrm{S}}_{\mathrm{B}}$

Therefore, A breaks before B

For ${\mathrm{r}}_{\mathrm{A}}=2{\mathrm{r}}_{\mathrm{B}}$

       ${\mathrm{S}}_{\mathrm{B}}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{\pi r}}_B^2} and T_A=\frac{{\mathrm{T}}_A}{{\mathrm{\pi r}}_{\mathrm{A}}^2}=\frac{4{\mathrm{T}}_{\mathrm{B}}}{\mathrm{\pi }{\left(2{\mathrm{r}}_{\mathrm{B}}\right)}^2}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{\pi r}}_{\mathrm{B}}^2}={\mathrm{S}}_{\mathrm{B}}$

As the stresses are equal, either of the wires may break.