JEE Main 2025 Apr 8 Shift 2, Physics Q3: Properties of Solids

JEE Main2025Apr 8, Shift 2Physics

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P \times 1 0^{11} Nm^{- 2}$ , where the value of $P$ is: (Take $g = 3 π$ m/s $^{2}$ )

A
$5$
B
$10$
C
$25$
D
$2.5$

Solution

Young's modulus relates stress and strain: $Y = \frac{F / A}{Δ ℓ / ℓ} = \frac{F ℓ}{A Δ ℓ}$ .

Substituting $F = m g = 50 \times 3 π = 150 π$ N, $A = π r^{2} = π (3 \times 1 0^{- 3})^{2} = 9 π \times 1 0^{- 6}$ m $^{2}$ , $ℓ = 3$ m and $Δ ℓ = 0.1 \times 1 0^{- 3}$ m.

$Y = \frac{150 π \times 3}{9 π \times 1 0 ^{- 6} \times 0.1 \times 1 0 ^{- 3}} = \frac{450}{9 \times 1 0 ^{- 10}} = 50 \times 1 0^{10} = 5 \times 1 0^{11}$ N/m $^{2}$ .

Comparing with $P \times 1 0^{11}$ gives $P = 5$ , matching option (1).

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JEE Main 2025 Apr 8 Shift 2, Physics Q3: Properties of Solids

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