Question

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young’s modulus is :

Answer 1

Answer:

Young's modulus of elasticity:

$\implies \sf Y = \dfrac{Longitudinal \: stress}{Longitudinal \: strain}$

• Longitudinal stress = F/A

• Longitudinal strain = ∆L/L

$\implies \sf Y = \dfrac{ \frac{F}{A} }{ \frac{\Delta L}{L} }$

$\implies \sf Y = \dfrac{ FL}{ A\Delta L }$

• Force acting on the body is 'mg'.

$\implies \sf Y = \dfrac{ MgL}{ A\Delta L}$

Initial Length is L and final Length is given as L₁.

$\implies \underline{ \boxed{ \red{ \bf Y = \dfrac{ MgL}{ A( L_{1} - L)}}}}$

Answer 2

Answer:

$\maltese \ \underline{\boxed{\mathrm{Y = \dfrac{MgL}{A(L_1-L)}}}}$

Step-by-step explanation:

Given :

A wire is hanging from a fixed support of

• length L
• area of cross section A

The length of the wire changes to L1 when mass M is suspended from its free end.

To find :

The expression for Young’s modulus

Solution :

We know that,

$\boxed{\mathrm{Y = \dfrac{Stress}{Strain}}}$

Stress = Force per Area and Strain is elongation per actual length

$\longmapsto {\mathrm{Y = \dfrac{\bigg(\dfrac{F}{A}\bigg)}{\bigg(\dfrac{e}{L}\bigg)}}}$

$\boxed{\mathrm{Y = \dfrac{FL}{Ae}}}$

where, F = force, Here, F = Mg as mass M is suspended from the free support

L = Length

A = Area

e = elongation, Here, e = L₁ - L

So, substituting these terms into the formula, we get,

$\bigstar \ \underline{\boxed{\mathbf{Y = \dfrac{MgL}{A(L_1-L)}}}}$

Hope it helps!!