The adjacent graph shows the extension (Δ l) of a wire of length 1 m suspended from the top of a root at one end with a load W connected to the other end. If the crosssectional area of the wire is 10-6 m 2, calculate the Young's modulus of the material 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 root at one end with a load W connected to the other end. If the crosssectional area of the wire is 10-6 m 2, calculate the Young's modulus of the material of the wire. (A) 2 × 1011 N / m 2 (B) 2 × 10-11 N / m 2 (C) 3 × 10-12 N / m 2 (D) 2 × 10-13 N / m 2

Tardigrade Admin · Accepted Answer

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

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

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

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

$3 \times 1 0^{- 12} N / m^{2}$

$2 \times 1 0^{- 13} N / m^{2}$

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

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

2×1011N/m2

2×10−11N/m2

3×10−12N/m2

2×10−13N/m2