A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface by a constant horizontal force F. The area of cross-section of the plank is A. the compressive strain on the plank in the direction of the force is

The correct option is F/AY.

Let us analyze the motion and deformation of the uniform plank step-by-step:

Step 1: Acceleration of the plank
Let the uniform plank have a mass $M$ and length $L$. The plank is moved over a smooth horizontal surface by a constant horizontal force $F$ applied at one end. Since the surface is smooth, there is no friction.
According to Newton's second law of motion, the acceleration $a$ of the plank is given by:
$a=\frac{F}{M}$

Step 2: Finding the internal force along the plank
Since the plank is accelerating, the internal force (compressive force) varies along its length. Let us consider a point at a distance $x$ from the end where the force $F$ is applied.
The portion of the plank of length $\left(L-x\right)$ behind this point has a mass:
$m=\frac{M}{L}\left(L-x\right)$
The force $F\left(x\right)$ at this section must accelerate this mass $m$ with the same acceleration $a$:
$F\left(x\right)=m\cdot a=\left[\frac{M}{L}\left(L-x\right)\right]\cdot \frac{F}{M}=F\left(1-\frac{x}{L}\right)$

Step 3: Compressive strain at the end where force is applied
The maximum compressive force and hence the maximum compressive strain occurs at the end where the force is applied, i.e., at $x=0$.
At $x=0$, the compressive force is:
$F\left(0\right)=F$
The compressive stress at this cross-section is:
$\text{Stress}=\frac{F}{A}$
Using Hooke's Law, Young's modulus $Y$ is defined as:
$Y=\frac{\text{Stress}}{\text{Strain}}$
Rearranging for the compressive strain, we get:
$\text{Strain}=\frac{\text{Stress}}{Y}=\frac{F}{AY}$