A uniform rod of length L and density `rho` is being pulled along a smooth floor with a horizontal acceleration `alpha` (see Fig.) The magnitude of the stress at the transverse cross-section through the mid-point of the rod is……..

Answer

A, B

Explanation

Let A be the area of cross-section of the rod.
Consider the back half portion of the rod.
Mass of half portion of the rod `=(rhoAL)/(2)`
The force responsible for its acceleration is
`f=(rhoAL)/(2)xxa` `:.` Stress `=f/A=(rhoLalpha)/(2)`