Question

The length of a wire is 3.0. m. The cross-sectional
area is 1.0 mm^2. How much work have to be done in
stretching its length by 0.2 mm ? Young's modulus of
the material of the wire is 2.0 x 10^11 newton/m^2.
Ans. 1.33 X 10^-3J.​

Answer 1

Given,

$\sf{L=3\ m}$

$\sf{A=10^{-6}\ m^2}$

$\sf{\Delta L=0.2\times10^{-3}\ m}$

$\sf{Y=2\times10^{11}\ N\ m^{-2}}$

If F is the force, the work done is,

$\longrightarrow\sf{W=\dfrac{1}{2}\cdot F\cdot \Delta L}$

$\longrightarrow\sf{W=\dfrac{1}{2}\cdot AY\cdot\dfrac{\Delta L}{L}\cdot\Delta L}$

$\longrightarrow\sf{W=\dfrac{AY(\Delta L)^2}{2L}}$

$\longrightarrow\sf{W=\dfrac{10^{-6}\times2\times10^{11}\times(0.2\times10^{-3})^2}{2\times3}}$

$\longrightarrow\sf{\underline{\underline{W=1.33\times10^{-3}\ J}}}$