JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Elasticity & SHM

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A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is 0.2 mm. If same force is applied to another steel wire 'B' of double the length and a diameter 2.4 times that of the wire 'A', the elongation in the wire 'B' will be (wires having uniform circular cross sections)

(A) $6.06 \times 1 0^{- 2} mm$
(B) $2.77 \times 1 0^{- 2} mm$
(C) $3.0 \times 1 0^{- 2} mm$
(D) $6.9 \times 1 0^{- 2} mm$ ✓ Correct

Answer: (D)

Solution

The elongation of a clamped wire under a force

F

Δ ℓ = \frac{F ℓ}{Y A}

, so for the same force and material

Δ ℓ \propto \frac{ℓ}{A} \propto \frac{ℓ}{d ^{2}}

. Wire

B

has twice the length and a diameter

2.4

times that of

A

, hence

\frac{Δ ℓ _{B}}{Δ ℓ _{A}} = \frac{2}{( 2.4 ) ^{2}} = \frac{2}{5.76}

. With

Δ ℓ_{A} = 0.2

mm,

Δ ℓ_{B} = 0.2 \times \frac{2}{5.76} = 0.069

= 6.9 \times 1 0^{- 2}

mm.

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Elasticity & SHM

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