Question

A force of 20N applied to the end of a wire 4m long produces an extension of 0.24 mm. If the diameter of wire is 2mm, calculate the stress on the wire, its strain and value of Young Modulus.​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

A force of 20N applied to the end of a wire 4m long produces an extension of 0.24 mm. If the diameter of wire is 2mm, calculate the stress on the wire, its strain and value of Young Modulus.

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Force (F) = 20N
• Wire lenght (L) = 4m
• Extension (∆L) = 0.24mm or ${\sf 0.24×10^{-3}m}$
• Diameter = 2mm therefore Radius (r) = 1mm or ${\sf 1×10^{-3}}$

$\large\underline{\underline{\sf To\:Find:}}$

• Stress
• Strain
• Young's Modulus

❏ Stress ➝

➝$\large{\boxed{\bf \blue{Stress=\dfrac{F}{A}}}}$

F = Force

A = Area = πr²

$\implies{\sf Stress=\dfrac{20}{π×(1×10^{-3})^2}}$

$\implies{\sf Stress=\dfrac{20}{π×10^{-6}} }$

$\implies{\bf \red{Stress=\dfrac{20×10^6}{π}N/m^2} }$

❏ Strain ➝

➝$\large{\boxed{\bf \blue{Strain=\dfrac{Change\:in\: Lenght}{Original\: Lenght} }}}$

$\implies{\sf Strain=\dfrac{4}{0.24×10^{-3}}}$

$\implies{\sf Strain = 0.06×10^{-3}}$

$\implies{\bf \red{Strain= 6×10^{-5}} }$

➝ It is dimensionless

❏ Young's Modulus ➝

➝$\large{\boxed{\bf \blue{Y=\dfrac{FL}{A\triangle L}}}}$

$\implies{\sf Y = \dfrac{20×4}{π×(10^{-3})^2×0.24×10^{-3}}}$

$\implies{\sf Y = \dfrac{80}{0.75×10^{-9}} }$

$\implies{\bf \red{Y = 1.06×10^7\:N/m^2} }$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Stress = ${\bf \red{Stress=\dfrac{20×10^6}{π}N/m^2} }$

Strain = ${\bf \red{Strain= 6×10^{-5}} }$

Young's Modulus = $\implies{\bf \red{Y = 1.06×10^7\:N/m^2} }$

Answer 2

Answer:

A force of 20N applied to the end of a wire 4m long produces an extension of 0.24 mm. If the diameter of the wire is 2mm. Then

1. Stress on the wire  =   $6.3694\times10^{6} N/ m^{2}$
2. Strain on the wire   =    $0.6 \times10^{-4} N/m^{2}$
3. Young's modulus    =    $10.6156\times10^{10} N/m^{2}$

Explanation:

We have the expression for Youg's modulus(Y) as,

$Y = \frac{Stress}{Strain}$

Stress = $\frac{Force( F )}{Area (A)}$      ... ( 1 )

 &  Strain = $\frac{\Delta L}{L}$        ... ( 2 )

That is,

$Y=\frac{F L}{A \Delta L}$                  ... ( 3 )

Where

F - The stretching force or tension on the wire.

L -  Initial length of the wire

A - Area

ΔL -  Extended length

Given,

F  =  20 N

L = 4 m

Diameter (d)  = 2 mm = 2 $\times 10^{-3} m$

Radius (r)  = $\frac{d}{2} = 1\times 10^{-3} m$

Thus,   A = $\pi r^{2} = 3.14\times( 1\times10^{-3})^{2} = 3.14 \times 10^{-6} m^{2}$

ΔL = 0.24 mm = $2.4\times 10^{-4} m$

• Stress on the wire

Substituting the values into the equation(1),

Equation (1) becomes,

$\frac{ F }{ A}$ =  $\frac{20}{ 3.14 \times 10^{-6} } = 6.3694\times10^{6} N/ m^{2}$

• Strain on the wire

Substituting values into equation(2)

$\frac{\Delta L}{L}$   =  $\frac{2.4\times 10^{-4} }{4}= 0.6 \times10^{-4} N/m^{2}$

• Young's modulus

Equation (3) becomes,

$Y = \frac{\ 6.3694\times10^{6} }{0.6\times10^{-4} } =10.6156\times10^{10} N/m^{2}$

 

Ans :

1. Stress on the wire  =   $6.3694\times10^{6} N/ m^{2}$
2. Strain on the wire   =    $0.6 \times10^{-4} N/m^{2}$
3. Young's modulus    =    $10.6156\times10^{10} N/m^{2}$