JEE Main 2026 — Physics Mechanics

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

The breaking stress for both wires is $\sigma_{max} = 12 \times 10^8 \text{ N/m}^2$σmax​=12×108 N/m2.

For the upper wire:

Area of cross-section, $A_u = 0.008 \text{ cm}^2 = 8 \times 10^{-7} \text{ m}^2$Au​=0.008 cm2=8×10−7 m2

Maximum tension the upper wire can withstand is:

$$T_{u,max} = \sigma_{max} \times A_u = (12 \times 10^8) \times (8 \times 10^{-7}) = 960 \text{ N}$$Tu,max​=σmax​×Au​=(12×108)×(8×10−7)=960 N

The total mass supported by the upper wire is $(30 + 10 + m) = (40 + m) \text{ kg}$(30+10+m)=(40+m) kg.

Tension in the upper wire, $T_u = (40 + m)g = (40 + m) \times 10$Tu​=(40+m)g=(40+m)×10

To avoid breaking the upper wire:

$$(40 + m) \times 10 \le 960$$(40+m)×10≤960

$$40 + m \le 96 \implies m \le 56 \text{ kg}$$40+m≤96⟹m≤56 kg

For the lower wire:

Area of cross-section, $A_l = 0.004 \text{ cm}^2 = 4 \times 10^{-7} \text{ m}^2$Al​=0.004 cm2=4×10−7 m2

Maximum tension the lower wire can withstand is:

$$T_{l,max} = \sigma_{max} \times A_l = (12 \times 10^8) \times (4 \times 10^{-7}) = 480 \text{ N}$$Tl,max​=σmax​×Al​=(12×108)×(4×10−7)=480 N

The total mass supported by the lower wire is $(10 + m) \text{ kg}$(10+m) kg.

Tension in the lower wire, $T_l = (10 + m)g = (10 + m) \times 10$Tl​=(10+m)g=(10+m)×10

To avoid breaking the lower wire:

$$(10 + m) \times 10 \le 480$$(10+m)×10≤480

$$10 + m \le 48 \implies m \le 38 \text{ kg}$$10+m≤48⟹m≤38 kg

To ensure neither wire breaks, the added mass $m$m must satisfy both conditions. Therefore, the maximum mass that can be added is the smaller of the two values:

$$m = 38 \text{ kg}$$m=38 kg

Answer: $38$38