In an experiment, brass and steel wires of length 1m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is : [Given, the Young's Modulus for steel and brass are respectively, 120 × 109 N/m2 and 60 × 109 N/m2]

Engineering

Physics

Stress Strain and Hookes Law New

Question

In an experiment, brass and steel wires of length 1m each with areas of cross section 1 mm² are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is :

[Given, the Young's Modulus for steel and brass are respectively, 120 × 10⁹ N/m² and 60 × 10⁹ N/m²]

8.0 × 10⁶ N/m²

1.2 × 10⁶ N/m²

1.8 × 10⁶ N/m²

0.2 × 10⁶ N/m²

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l = 1 M

A = 10^–6 M²

$Stress = \frac{F}{A}$

$Stress = \frac{Stress}{Y}$

$Δℓ = \frac{ℓ \times F}{AY}$

${Δℓ}_{1} + {Δℓ}_{2} = \frac{ℓ_{1} \times F}{{AY}_{1}} + \frac{ℓ_{2} \times F}{{AY}_{2}} = 0 .2 \times 10^{- 3}$

$\frac{F}{A} = \frac{0 .2 \times 10^{- 3}}{\frac{ℓ}{Y_{1}} + \frac{ℓ}{Y_{2}}}$

$= \frac{0 .2 \times 10^{- 3}}{\frac{1}{120 \times 10^{9}} + \frac{1}{60 \times 10^{9}}} = \frac{0 .2 \times 10^{- 3} \times 10^{9} \times 120}{1 + 2}$

$= \frac{0 .2 \times 10^{6} \times 120}{3} = 8 \times 10^{6}$