Determination of Young’s Modulus of the Material of a Wire

A typical experimental arrangement to determine the Young’s modulus of a material of wire under tension is shown in Fig. 1. It consists of two long straight wires of same length and equal radius suspended side by side from a fixed rigid support. The wire A (called the reference wire) carries a millimetre main scale M and a pan to place a weight. 

Fig. 1 An arrangement for the determination of Young’s modulus of the material of a wire.

The wire B (called the experimental wire) of uniform area of cross-section also carries a pan in which known weights can be placed. A vernier scale V is attached to a pointer at the bottom of the experimental wire B, and the main scale M is fixed to the reference wire A. The weights placed in the pan exert a downward force and stretch the experimental wire under a tensile stress. The elongation of the wire (increase in length) is measured by the vernier arrangement. 

The reference wire is used to compensate for any change in length that may occur due to change in room temperature, since any change in length of the reference wire due to temperature change will be accompanied by an equal change in experimental wire.

Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted. Now the experimental wire is gradually loaded with more weights to bring it under a tensile stress and the vernier reading is noted again. 

The difference between two vernier readings gives the elongation produced in the wire. Let r and L be the initial radius and length of the experimental wire, respectively. Then the area of cross-section of the wire would be $\pi r^2$ . Let M be the mass that produced an elongation ∆L in the wire. Thus the applied force is equal to Mg, where g is the acceleration due to gravity. 

From Eq. Y = (F/A)/(∆L/L) = (F × L)/(A × ∆L), the Young’s modulus of the material of the experimental wire is given by

$Y=\frac{\sigma}{\epsilon}=\frac{mg}{\pi r^2}\frac{L}{\Delta L}$