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 cross-sectional area of the wire is 10-6 m2 , 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 roof at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10-6 m2 , calculate the Young's modulus of the material of the wire. (A) 2 × 1011 Nm-2 (B) 2 × 10-11 Nm-2 (C) 3 × 10-12 Nm-2 (D) 2 × 10-13 Nm-2

Tardigrade Admin · Accepted Answer

From the graph l = 10-4m ,F = 20 N A = 10-6 m2 , L = 1 m ∴ Y = (FL/Al) = (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 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 1 m suspended from the top of a roof 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.

2×1011Nm−2

2×10−11Nm−2

3×10−12Nm−2

2×10−13Nm−2