A uniform cylindrical rod of length ‘L, area of cross-section ‘A’ and Young’s modulus ‘Y’ is acted upon by the forces as shown in the figure. The elongation of the rod is

The elongation of the rod can be calculated using the equation for stress-strain relationship:

$\mathrm{Stress} =\frac{ \mathrm{Force}}{\mathrm{Area}}= \mathrm{Y} \times \mathrm{Strain}$

where Y is Young's modulus, Stress is the force per unit area, and Strain is the fractional change in length of the rod.

Let the elongation of the rod be ΔL. Then, the strain can be calculated as:

$\mathrm{Strain} = \frac{\mathrm{\Delta L}}{\mathrm{L}}$

Combining these equations, we get:

$\frac{\mathrm{\Delta L}}{\mathrm{L}}= \frac{\mathrm{Stress}}{\mathrm{Y}}= \frac{\left({\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}\right)}{\mathrm{Y}} = \frac{\mathrm{F}}{\mathrm{A}\times \mathrm{Y}}$

where F is the net force acting on the rod. So, the elongation of the rod can be expressed as:

From the figure,

$\Delta l=\Delta l_1+\Delta l_2=\frac{3\mathrm{F}\left({\displaystyle \frac{2\mathrm{L}}{3}}\right)}{\mathrm{AY}}+\frac{2\mathrm{F}\left({\displaystyle \frac{\mathrm{L}}{3}}\right)}{\mathrm{AY}}=\frac{8\mathrm{FL}}{3\mathrm{AY}}$

Hence the correct answer is $\frac{8\mathrm{FL}}{3\mathrm{AY}}.$