JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Elasticity & SHM

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) $2$ ✓ Correct
(C) $1.9$
(D) $1$

Answer: (B)

Solution

Each string carries the full load

m g

. For string A,

Δ ℓ_{A} = \frac{m g ℓ}{A Y}

; for string B of length

2 ℓ

and modulus

2 Y

Δ ℓ_{B} = \frac{m g ( 2 ℓ )}{A ( 2 Y )} = \frac{m g ℓ}{A Y}

, equal to that of A. The total is

Δ ℓ = \frac{2 m g ℓ}{A Y}

. With

m = 0.8

kg,

g = 10

m/s

^{2}

ℓ = 0.314

Y = 2 \times 1 0^{10}

N/m

^{2}

and

A = π r^{2} = π (0.2 \times 1 0^{- 3})^{2}

^{2}

Δ ℓ = \frac{2 ( 0.8 ) ( 10 ) ( 0.314 )}{π ( 4 \times 1 0 ^{- 8} ) ( 2 \times 1 0 ^{10} )} = 2 \times 1 0^{- 3}

= 2

mm.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Elasticity & SHM

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