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Physics Question 43 – JEE-MAIN 2026

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Two wires as shown in the figure below, made of steel and have breaking stress of 12×108 N/m2. Area of cross-section of upper wire is 0.008 cm2 and of lower wire is 0.004 cm2. The maximum mass that can be added to pan without breaking any wire is _______ kg. (take g=10 m/s2)

Identify the total downward force (weight) acting on each wire due to the masses hanging below it.

Step 1: Calculate Breaking Force for Each Wire✦ Active

First, convert the cross-sectional areas from cm2 to m2 and then calculate the maximum force each wire can withstand using the breaking stress formula Fbreak=σbreak×A.

Au=0.008 cm2=0.008×10−4 m2=8×10−7 m2 Al=0.004 cm2=0.004×10−4 m2=4×10−7 m2 σbreak=12×108 N/m2 Fu,max=(12×108 N/m2)×(8×10−7 m2)=960 N Fl,max=(12×108 N/m2)×(4×10−7 m2)=480 N
Step 2: Determine Tension in Each Wire with Additional Mass○ Expand

Let M be the additional mass added to the pan. Calculate the tension in the lower wire (Tl) and the upper wire (Tu) due to the hanging masses and the additional mass M, taking g=10 m/s2.

Tl=(10 kg+M)g=(10+M)×10=100+10M N Tu=(30 kg+10 kg+M)g=(40+M)×10=400+10M N
Step 3: Apply Breaking Conditions and Find Maximum M○ Expand

For each wire, the tension must not exceed its maximum breaking force. Set up inequalities for M for both wires and find the maximum M that satisfies both conditions.

For lower wire: Tl≤Fl,max⟹100+10M≤480⟹10M≤380⟹M≤38 kg For upper wire: Tu≤Fu,max⟹400+10M≤960⟹10M≤560⟹M≤56 kg For both wires to remain intact, M must satisfy both conditions. Mmax=min(38 kg,56 kg)=38 kg
💡 Teacher's Secret Hint

The system will break at the weakest point. Therefore, the maximum additional mass is limited by the wire that reaches its breaking point first.

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