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 cross-sectional 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 cross-sectional area of the wire is 10-6 m 2, calculate the Young's modulus of the material of the wire. (A)  2× 1011 N/m2  (B)  2× 10-11N/m2  (C)  3× 10-12 N/m2  (D)  2× 10-13 N/m2

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 cross-sectional 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 cross-sectional 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