JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

A string $A$ of length $0.314$ m of Young's modulus $2 \times 1 0^{10}$ N/m $^{2}$ is connected to another string $B$ of length and Young's modulus both twice of those of $A$ . This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass $0.8$ kg. The net change in length of the combination is _______ mm.

(radius of both the strings is $0.2$ mm and acceleration due to gravity $= 10$ m/s $^{2}$ )

(Mass of both strings is to be neglected as compared to the mass of load)

A
$3$
B
$2$
C
$1.9$
D
$1$

Solution

Both strings carry the same tension $F = m g$ . Same cross-sectional area $A = π r^{2}$ .

String A: length $ℓ$ , Young's modulus $Y$ . $Δ ℓ_{A} = \frac{F ℓ}{A Y} .$

String B: length $2 ℓ$ , Young's modulus $2 Y$ . $Δ ℓ_{B} = \frac{F \cdot 2 ℓ}{A \cdot 2 Y} = \frac{F ℓ}{A Y} .$

Total: $Δ ℓ = Δ ℓ_{A} + Δ ℓ_{B} = \frac{2 F ℓ}{A Y} = \frac{2 m g ℓ}{A Y} .$

Numerics: $A = π (0.2 \times 1 0^{- 3})^{2} = 3.14 \times 4 \times 1 0^{- 8}$ m $^{2} = 1.256 \times 1 0^{- 7}$ m $^{2}$ .

$Δ ℓ = \frac{2 \times 0.8 \times 10 \times 0.314}{1.256 \times 1 0 ^{- 7} \times 2 \times 1 0 ^{10}} = \frac{5.024}{2512} \approx 2 \times 1 0^{- 3}$ m $= 2$ mm.

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