JEE Main 2026 Apr 6 Shift 2, Physics Q8: Properties of Solids

JEE Main2026Apr 6, Shift 2Physics

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 $1 0^{- 5}$ m $^{2}$ 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
$1.0 \times 1 0^{10}$
D
$2.0 \times 1 0^{11}$

Solution

From the graph, the slope is $\frac{Δ l}{W} = \frac{6 \times 1 0 ^{- 4}}{60} = 1 0^{- 5}$ m/N.

Young's modulus: $Y = \frac{W / A}{Δ l / L} = \frac{W L}{A Δ l} = \frac{L}{A} \cdot \frac{W}{Δ l}$

$Y = \frac{1}{1 0 ^{- 5}} \cdot \frac{1}{1 0 ^{- 5}} = 1 0^{10}$ N/m $^{2}$ .

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JEE Main 2026 Apr 6 Shift 2, Physics Q8: Properties of Solids

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