Question

A rod of mass M is placed over smooth horizontal surface. Forces 2F, F are applied at the ends of the rod as shown. If length, cross-sectional area and Young's modulus of rod are L, A and Y respectively, then total elongation of rod will be (F=Mg)


e

• A. $\frac{4MgL}{YA}$
• B. $\frac{MgL}{2YA}$
• C. $\frac{MgL}{YA}$
• D. $\frac{3MgL}{2YA}$

Answer 1

The correct option is D $\frac{3MgL}{2YA}$

$dI=\frac{T}{\left(\frac{YA}{dx}\right)}dI=\frac{T}{YA}dxdI=\left(\frac{Mg}{YA}\right)(2-\frac{x}{L})dx\Delta I={\int }_0^L\left(\frac{Mg}{YA}\right)(2-\frac{x}{L})dx$

$\Delta I=\frac{3}{2}\frac{MgL}{YA}$