Question bank/JEE Main/Physics

A uniform elastic rod of cross-section area A, natural length L
and Young's modulus Y is placed on asmooth horizontal surface. Now two
horizontal forces (of magnitude F and 3F) directed along the lengthof
rod and in opposite direction act at two of its ends as shown. After the
rod has acquired steady state,the extension of the rod will be.

A$\frac{2F}{YA}L$B$\frac{4F}{YA}L$C$\frac{F}{YA}L$D$\frac{3F}{2YA}L$

Show answer and solution▾

Tension in rod at a distance x from right edge is
$T = F\left( 3 - 2\frac{x}{L} \right)$
[IMAGE] $\therefore$net extension in
rod $= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx