Question

5)Young's modulus of the material of a wire is 9.68 x 10^10 N/m^2. A wire of material of diameter 0.95 mm is stretched by applying a certain force. What should be the limit of this force, if the strain is not to exceed 1 in 1000?
(F = 68.58 N)​

Answer 1

Answer:

68.6 MN

Explanation:

strain ∈ = 10∧-3

stress = E∈ =  9.68 x 10^10 N/m^2x 10∧-3  = 9.68 x 10^7 N/m^2

Force = 9.68 x 10^7 N/m^2 x π ( 475)²×10∧-6 mm² = 68613950N = 68.6 MN

Answer 2

Concept:

The ratio of tensile stress to tensile strain is known as Young's modulus (E), a feature of the material that indicates how easily it can stretch and flex.

Up to an elastic limit, there is a directly proportional relationship between stress and strain.

A material's ability to regain its former shape after being stretched or squeezed under certain stress.

Given:

The Youngs modulus of the material is $9.68\times10^{10}N/m^2$.

The diameter of the wire is $0.95mm$.

The strain does not exceed 1 in 1000.

Find:

The stretching force.

Solution:

The Youngs modulus of a material or a wire is given by:

Y= Stress/ Strain

Now,

Strain $=\frac{1}{1000}=10^-3$

The radius of the wire $=\frac{0.95}{2}=0.475mm=4.75\times10^{-4}$

Now,

$Y=\frac{\frac{F}{A} }{10^{-3}}$

$9.68\times10^{10}=\frac{F}{\pi r^2\times10^{-3}}$

$F=9.68\times10^{10}\times(4.75\times10^{-4})\times3.14\\F=68.62N$

The limiting stretching force is $68.62N$.

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