The graph shown the extension of is wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross sectional area of the wire is 1 mm 2, then the Young's modulus of the material of the wire.

Question

The graph shown the extension of is wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross sectional area of the wire is 1 mm 2, then the Young's modulus of the material of the wire. (A) 2 × 1011 Nm -1 (B) 2 × 1010 Nm -2 (C) 1 / 2 × 1011 Nm -2 (D) None of these

Tardigrade Admin · Accepted Answer

Y =( F / A /Δ ℓ / ℓ)=( W ℓ/ A Δ ℓ) ⇒ ( W /Δ ℓ)=( YA /ℓ)= slope ⇒ Y =(ℓ/ A )( slope ) =(1/10-6)((40-20/(2-1) × 10-3)) =2 × 1010 Nm -2

Q. The graph shown the extension of is wire of length $1 m$ suspended from the top of a roof at one end and with a load $W$ connected to the other end. If the cross sectional area of the wire is $1 m m^{2}$ , then the Young's modulus of the material of the wire.

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

$2 \times 1 0^{10} N m^{- 2}$

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

None of these

$Y = \frac{F / A}{Δ ℓ / ℓ} = \frac{W ℓ}{A Δ ℓ}$
$\Rightarrow \frac{W}{Δ ℓ} = \frac{Y A}{ℓ} =$ slope
$\Rightarrow Y = \frac{ℓ}{A} ($ slope $)$
$= \frac{1}{1 0 ^{- 6}} (\frac{40 - 20}{( 2 - 1 ) \times 1 0 ^{- 3}})$
$= 2 \times 1 0^{10} N m^{- 2}$

Q. The graph shown the extension of is wire of length 1m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross sectional area of the wire is 1mm2, then the Young's modulus of the material of the wire.

2×1011Nm−1

2×1010Nm−2

1/2×1011Nm−2