JEE Main 2026 — Mechanical Properties of Solids Question with Solution

Let the maximum mass that can be added to the pan be $m$.

The breaking stress for both wires is ${\sigma }_{max}=12\times 10^8{\text{ N/m}}^2$.

For the upper wire:
Area of cross-section, $A_u=0.008{\text{ cm}}^2=8\times 10^{-7}{\text{ m}}^2$
Maximum tension the upper wire can withstand is:
$T_{u,max}={\sigma }_{max}\times A_u=(12\times 10^8)\times (8\times 10^{-7})=960\text{ N}$
The total mass supported by the upper wire is $(30+10+m)=(40+m)\text{ kg}$.
Tension in the upper wire, $T_u=(40+m)g=(40+m)\times 10$
To avoid breaking the upper wire:
$(40+m)\times 10\le 960$
$40+m\le 96⟹m\le 56\text{ kg}$

For the lower wire:
Area of cross-section, $A_l=0.004{\text{ cm}}^2=4\times 10^{-7}{\text{ m}}^2$
Maximum tension the lower wire can withstand is:
$T_{l,max}={\sigma }_{max}\times A_l=(12\times 10^8)\times (4\times 10^{-7})=480\text{ N}$
The total mass supported by the lower wire is $(10+m)\text{ kg}$.
Tension in the lower wire, $T_l=(10+m)g=(10+m)\times 10$
To avoid breaking the lower wire:
$(10+m)\times 10\le 480$
$10+m\le 48⟹m\le 38\text{ kg}$

To ensure neither wire breaks, the added mass $m$ must satisfy both conditions. Therefore, the maximum mass that can be added is the smaller of the two values:
$m=38\text{ kg}$

Answer: $38$