JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 1 0^{8}$ N/m $^{2}$ . Area of cross-section of upper wire is $0.008$ cm $^{2}$ and of lower wire is $0.004$ cm $^{2}$ . The maximum mass that can be added to pan without breaking any wire is _______ kg. (take $g = 10$ m/s $^{2}$ )

A
56
B
38
C
96
D
5.6

Solution

Let $m$ be the extra mass on the pan. The lower wire supports $(10 + m)$ kg; the upper wire supports $(40 + m)$ kg.

Upper-wire stress limit: $\frac{( 40 + m ) \cdot 10}{0.008 \times 1 0 ^{- 4}} \leq 12 \times 1 0^{8}$ gives $40 + m \leq 96$ , so $m \leq 56$ kg.

Lower-wire stress limit: $\frac{( 10 + m ) \cdot 10}{0.004 \times 1 0 ^{- 4}} \leq 12 \times 1 0^{8}$ gives $10 + m \leq 48$ , so $m \leq 38$ kg.

The binding constraint is the lower wire. The maximum safe mass is $38$ kg.

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