JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Elasticity & SHM

A wire breaks when its tensile stress reaches $12\times 10^8{\text{N/m}}^2$. From the figure the upper wire ($A=0.008{\text{cm}}^2=8\times 10^{-7}{\text{m}}^2$) carries the $30$ kg block, the $10$ kg block, the pan and the added mass $m$, a total of $(40+m)$ kg, while the lower wire ($A=0.004{\text{cm}}^2=4\times 10^{-7}{\text{m}}^2$) carries only the $10$ kg block, the pan and $m$, i.e. $(10+m)$ kg. For the upper wire, $\frac{(40+m)\times 10}{8\times 10^{-7}}=12\times 10^8$ gives $40+m=96$, so $m=56$ kg. For the lower wire, $\frac{(10+m)\times 10}{4\times 10^{-7}}=12\times 10^8$ gives $10+m=48$, so $m=38$ kg. The wire reaching its breaking stress first fixes the limit, so the maximum mass that can be added to the pan is $38$ kg.