Question

In an experiment, brass and steel wires of length 1 m 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 requires to produced a new 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]
(A) 1.8 × 10⁶ N/m²
(B) 0.2 × 10⁶ N/m²
(C) 1.2 × 10⁶ N/m²
(D) 4.0 × 10⁶ N/m²

Answer 1

The stress required to produced a new elongation of 0.2 mm is 8 x $10^{6}$ N/$m^{2}$.

Given Young’s Modulus for

• steel , Y1 =  120 × 10⁹ N/m² and

• brass Y2 = 60 × 10⁹ N/m²
• length of both the wires = 1m

• k1 = Y1 A1/ l1 =  120 × 10⁹ N/m² A / 1

• k2 = Y2 A2/ l2 =  60 × 10⁹ A /1

Since the wires are connected in series,

• Keq = K1 .K2/(K1 + K2)  
• Keq =  120 x 60 /( 120 + 60) × 10⁹ A N/m

• Keq = 40 × 10⁹ A N/m

Therefore ,

Stress required to produce elongation of ΔL = 0.2mm is,

Stress = Force/ Area = F/A

F = Keq x ΔL    = 40 × 10⁹ A x ΔL = (40 × 10⁹)( $10^{-6}$)(0.2 x $10^{-3}$ ) = 8 N

• F/ A = 8 x $10^{6}$ N/$m^{2}$