A steel wire of density $8000\;kg/m^3$ weights $24\;gm$ and is $250\;cm$ long. It lengthens by $1.2\;mm$ when stretched by a force of $80\;N$. Calculate the Young's modulus for the steel and the energy stored in the wire.

Given,
Density of steel $(\rho) = 8000\;kg/m^3$
Mass of the steel$(m) = 24\;gm = 24 * 10^{-3}\;kg$
Length of wire $(l) = 2.5\;m$
Elongation $(e) = 1.2\;mm = 1.2* 10^{-3}\;m$
Force $(F) = 80\;N$
Young's modulus of Elasticity $(Y) = ?$
Energy stored $(E) = ?$

We have,

$Y = \frac{F/A}{e/l} = \frac{F\,.\,l}{A\,.\,e}$ .......... (i)

But,

$\rho = \frac{m}{V} = \frac{m}{A\,.\,l}$   ⇒ $A = \frac{m}{\rho \,.\,l}$

Then substituting the  value of $A$ in equation $(i)$, we get

$Y = $ $\frac{F\,.\,l^2\,.\,\rho}{e\,.\,m} = \frac{80 \, * \, 2.5^2 \, * \, 8000}{1.2 \, * \, 10^{-3}\, * \, 24\,*\,10^{-3}}$ $ = 1.4\,*\,10^{11}\;N/m^2$

Again,

$E = \frac{1}{2} \,F\,.\,e = \frac{1}{2}\,80 \,*\,1.2 \,*\,10^{-3}$ $ = 4.8 \, * \, 10^{-2}\;J$

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