JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Elasticity & SHM

Figure represents the extension (

Δ l

) of a wire of length 1 meter, suspended from the ceiling of the room at one end with load W connected to the other end. If the cross-sectional area of the wire is

1 0^{- 5}

^{2}

then the Young's modulus of the wire is ______ N/m

^{2}

(A) $1.0 \times 1 0^{11}$
(B) $2.0 \times 1 0^{10}$
(C) $1.0 \times 1 0^{10}$ ✓ Correct
(D) $2.0 \times 1 0^{11}$

Answer: (C)

Solution

The straight line through the origin has slope

\frac{Δ l}{W} = \frac{1 \times 1 0 ^{- 4} m}{10 N} = 1 0^{- 5} m/N

. Young's modulus is

Y = \frac{W / A}{Δ l / l} = \frac{W}{Δ l} \cdot \frac{l}{A} = \frac{1}{1 0 ^{- 5}} \times \frac{1}{1 0 ^{- 5}} = 1.0 \times 1 0^{10} N/m^{2}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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