Class 11 Physics Mechanical Properties of Solids Determine Youngs Modulus

Determination of Young’s Modulus of the material of the wire

Experimental set up:-

• Two strings were hung from a support and two pans were attached to both the strings.
• Weights are kept on both the pans.
• When the number of weights in second pan was increased, the string got stretched and moved in downward direction.
• The change in length was measured by the metre scale which was kept on reference wire.
• Using this experiment, the Young’s modulus value was calculated
• Y= longitudinal stress/ longitudinal strain = σ/ ε
• = (F/A)/ (ΔL/L)
• Where original length = L and ΔL = change in length, F=mg (acting downwards) and A (area of cross-section of wire) = πr2
• = (mg/ πr2)/ (ΔL/L)
• Y= mgL/ πr2 ΔL
• This is the way to calculate the Young’s modulus.

Problem:- Read the following two statements below carefully and state, with reasons, if it is true or false.

1. a) The Young’s modulus of rubber is greater than that of steel;
2. b) The stretching of a coil is determined by its shear modulus.

(a) False

(b) True

For a given stress, the strain in rubber is more than it is in steel.

Young’s modulus, Y = 𝑆𝑡𝑟𝑒𝑠𝑠/𝑆𝑡𝑟𝑎𝑖𝑛

For a constant stress:  𝑌 ∝ 1/𝑆𝑡𝑟𝑎𝑖𝑛

Hence, Young’s modulus for rubber is less than it is for steel.

Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

Problem:- A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

Answer:- The tension force acting on each wire is the same.

Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.

The relation for Young’s modulus is given as:

Y =𝑆𝑡𝑟𝑒𝑠𝑠/𝑆𝑡𝑟𝑎𝑖𝑛 = (F/A)/Strain = (4F/ πd2)/Strain                Equation (i)

Where,

F = Tension force

A = Area of cross-section

D=Diameter of the wire

From equation (i) Y ∝ 1/d2

Young’s Modulus for iron, Y1 =190x109 Pa

Diameter of the iron wire =d1

Young’s Modulus for copper, Y2 = 110x109 Pa

Diameter of the copper wire =d2

Therefore the ratios of their diameters are given as:

d1/ d2 =√ Y1/ Y2 =√190x109/110x109 = √19/11 =1.31:1

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