A steel wire of length 4.7 m and cross-sectional area 3.0...

A steel wire of length $4.7 \mathrm{~m}$ and cross-sectional area $3.0 \times 10^{-5} \mathrm{~m}^{2}$ stretches by the same amount as a copper wire of length $3.5 \mathrm{~m}$ and cross-sectional area of $4.0 \times 10^{-5} \mathrm{~m}^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Key Concepts

Stress and Strain

Stress is the force applied per unit area, and strain is the ratio of the change in length to the original length. These concepts are central to understanding elastic deformation since their ratio, provided the deformation is within the elastic limit, gives the Young's modulus, a key indicator of a material's resistance to deformation.

Geometrical Factors in Elastic Deformation

The deformation of a material under load also depends on its dimensions, notably its length and cross-sectional area. A longer specimen or one with a smaller cross-sectional area will deform more for a given load. Understanding how these geometrical factors interplay with material properties is crucial for comparing the elastic response of different objects.

Hooke's Law

Hooke's Law states that, within the elastic limit of a material, the extension or compression is directly proportional to the applied force. This linear relationship, expressed neatly by the formula F = (YA/L), connects the force, the material's Young's modulus, and its geometric dimensions (length and cross-sectional area), allowing predictions of deformation.

Young's Modulus

Young's modulus is a fundamental material property that measures stiffness, defined as the ratio of stress (force per unit area) to strain (relative deformation). It quantifies how much a material will stretch or compress under a given load, forming the basis for comparing the elastic behavior of different materials.

Transcript

00:01 This question belongs to the mechanical properties of solid in which we have given a steel wire.

00:05 We have some given data.

00:07 Steel wire of length.

00:08 So l1 this is equal to 4 .7 meter and cross -sectional area a1, this is equal to 3 .0 multiplier by 10 to the power minus 5 meter square.

00:18 And it is stressed by the same amount as copper wire of length, l2, this is equals to 3 .5 meter and cross -sectional area, a2, equals to 4 .5.

00:30 4 .0 multiplied 10 to the power minus 5 meter square under the given road.

00:35 So that is delta l and force f both is same, okay, for copper wire and steel wire.

00:43 So we have to determine the ratio of the youngs modulus of the steel that is y1 to the y2, okay? that is copper, okay? so we know that the youngs modulus is given by y it is equals to force divided by area that is stress divided by strain that is delta l by capital l okay so from here we can see that this relation can be written as f multiplied by l divided by delta l multiplied by area a now we can see here that f and delta l both are same for both material so hence we can write here that young's modulus y it is directly proportional to l by area a so we can write this equation that y1 by y2, y1 by y2, this will be equals to l1 by a1 multiplied by a2 by l2.

01:36 So now substituting values, so we will get that l1 which is 4 .7 meter multiplied by a2 which is this 4 .0 modiplat be 10 to the power minus 5 meter square divided by area a1 which is 3 .0 multiplied by length l2 which is 3 .5 meter...

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