The force required to stretch a wire of cross-section $1 \mathrm{~cm}^{2}$ to double its length will be : (Given Young's modulus of the wire $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$)

(C) $2 \times 10^{7} \mathrm{~N}$

(A) $1 \times 10^{7} \mathrm{~N}$

(B) $1.5 \times 10^{7} \mathrm{~N}$

(D) $2.5 \times 10^{7} \mathrm{~N}$

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Elasticity & SHM

The force required to stretch a wire of cross-section

1 cm^{2}

to double its length will be :
(Given Young's modulus of the wire

= 2 \times 1 0^{11} N / m^{2}

)

Answer: (C)

Young's modulus:

Y = \frac{F / A}{Δ ℓ / ℓ}

, so

F = Y A \frac{Δ ℓ}{ℓ}

. Doubling the length means

Δ ℓ = ℓ

, so

\frac{Δ ℓ}{ℓ} = 1

. Given

Y = 2 \times 1 0^{11}

N/m² and

A = 1

cm²

= 1 0^{- 4}

m², we have

F = 2 \times 1 0^{11} \times 1 0^{- 4} \times 1 = 2 \times 1 0^{7}

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option