In determination of young's modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is '1 m'. The curve between extension and stress is depicted then Young's modulus of wire will be :-

Question

In determination of young's modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is '1 m'. The curve between extension and stress is depicted then Young's modulus of wire will be :- (A) 2 × 109 N / m 2 (B) 1 × 109 N / m 2 (C) 2 × 1010 N / m 2 (D) 1 × 1010 N / m 2

Tardigrade Admin · Accepted Answer

Slope =(Δ L / F / A ) =(7 × 10-3 m /7000 × 103 N / m 2) Δ L =( FL / dot A Y ) ⇒ (1/ Y )=(Δ L / overline F / A L ) =(7 × 10-3/7000 × 10-3 × 1) m 2 / N On solving Y =1 × 109 N / m 2

Q. In determination of young's modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is ' $1 m$ '. The curve between extension and stress is depicted then Young's modulus of wire will be :-

$2 \times 1 0^{9} N / m^{2}$

$1 \times 1 0^{9} N / m^{2}$

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

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

Slope $= \frac{Δ L}{F / A}$
$= \frac{7 \times 1 0 ^{- 3} m}{7000 \times 1 0 ^{3} N / m ^{2}}$
$Δ L = \frac{F L}{A ˙ Y}$
$\Rightarrow \frac{1}{Y} = \frac{Δ L}{F / A L}$
$= \frac{7 \times 1 0 ^{- 3}}{7000 \times 1 0 ^{- 3} \times 1} m^{2} / N$
On solving $Y = 1 \times 1 0^{9} N / m^{2}$

Q. In determination of young's modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is '1m'. The curve between extension and stress is depicted then Young's modulus of wire will be :-

2×109N/m2

1×109N/m2

2×1010N/m2

1×1010N/m2