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A Steel Wire and a Copper Wire of Equal Length and Equal Cross-sectional Area Are Joined End to End and the Combination is Subjected to a Tension. Find Ratio of the Stresses Developed in Two Wires . - Physics

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Short Note

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in the two wires .

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Solution

Given:
Young's modulus of steel = 2 × 1011 N m−2
Young's modulus of copper = 1.3 × 10 11 N m−2
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
 As per the question :

\[L_{\text{ steel}} = L_{\text{Cu}} \]
\[ A_{\text{ steel}} = A_{\text{Cu}} \] 
\[ F_{\text{Cu}} = F_{\text{Steel }} \]

Here: Lsteel and LCu denote the lengths of steel and copper wires, respectively.
           Asteel and ACu denote the cross-sectional areas of steel and copper wires, respectively.
           Fsteel and FCu  denote the tension of steel and cooper wires, respectively. 

\[\frac{\text{ Stress of Cu}}{\text{ Stress of Steel }} = \frac{F_{\text{Cu}}}{A_{\text{Cu }}}\frac{A_{\text{ Steel }}}{F_{\text{ Steel}}} = 1\]
Concept: Elastic Moduli - Determination of Young’s Modulus of the Material of a Wire
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 14 Some Mechanical Properties of Matter
Q 4.1 | Page 300
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