The adjacent graph shows the extension (Δ ℓ) of a wire of length 1 m suspended from the top of a roof at one end with a load w connected to the other end. If the crosssectional area of the wire is 10-6 m2, what is Young’s modulus of the wire?

Question

The adjacent graph shows the extension (Δ ℓ) of a wire of length 1 m suspended from the top of a roof at one end with a load w connected to the other end. If the crosssectional area of the wire is 10-6 m2, what is Young’s modulus of the wire? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b377dee23f9b8a962be1688ef16e3922-.png" /> (A) 2 × 1011 Nm -2 (B) 2 × 10-11 Nm -2 (C) × 10-12 Nm -2 (D) 2 × 10-14 Nm -2

Tardigrade Admin · Accepted Answer

From the graph, ℓ=10-4 m , F =20 N ℓ=1 × 10-4 m Given: A =10-6 m 2, L =1 m ∴ Y=(F L/A ℓ) =(20 × 1/10-6 × 10-4) =20 × 1010 =2 × 1011 Nm -2

Q. The adjacent graph shows the extension $(Δ ℓ)$ of a wire of length 1 m suspended from the top of a roof at one end with a load w connected to the other end. If the crosssectional area of the wire is $1 0^{- 6} m^{2}$ , what is Young’s modulus of the wire?

$2 \times 1 0^{11} N m^{- 2}$

$2 \times 1 0^{- 11} N m^{- 2}$

$\times 1 0^{- 12} N m^{- 2}$

$2 \times 1 0^{- 14} N m^{- 2}$

From the graph, $ℓ = 1 0^{- 4} m, F = 20 N$
$ℓ = 1 \times 1 0^{- 4} m$
Given: $A = 1 0^{- 6} m^{2}, L = 1 m$
$∴ Y = \frac{F L}{A ℓ}$
$= \frac{20 \times 1}{1 0 ^{- 6} \times 1 0 ^{- 4}}$
$= 20 \times 1 0^{10}$
$= 2 \times 1 0^{11} N m^{- 2}$

Q. The adjacent graph shows the extension (Δℓ) of a wire of length 1 m suspended from the top of a roof at one end with a load w connected to the other end. If the crosssectional area of the wire is 10−6m2, what is Young’s modulus of the wire?

2×1011Nm−2

2×10−11Nm−2

×10−12Nm−2

2×10−14Nm−2