A uniform rod of length L has a mass per unit length λ and area of cross section A. The elongation in the rod is l due to its own weight if it is suspended from the ceiling of a room. The Young’s modulus of the rod is

The correct option is λgL²/2Al.

Step-by-step Derivation:

Consider a uniform rod of length $L$, cross-sectional area $A$, and mass per unit length $\lambda$, suspended vertically from a ceiling.

To find the total elongation due to its own weight, let us analyze the tension at a distance $y$ from the bottom (free) end of the rod.
The mass of the section of the rod below this point of length $y$ is given by:
$m=\lambda y$

Therefore, the tension $T\left(y\right)$ in the rod at a distance $y$ from the bottom end is equal to the weight of this lower portion:
$T\left(y\right)=mg=\lambda yg$

Now, let us consider an infinitesimal element of length $dy$ at this position. According to the definition of Young's modulus $Y$, stress is proportional to strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T\left(y\right)/A}{dl/dy}$

Rearranging this expression to solve for the small elongation $dl$ of the element:
$dl=\frac{T\left(y\right)dy}{AY}=\frac{\lambda gydy}{AY}$

To find the total elongation $l$ of the rod, we integrate this expression from $y=0$ to $y=L$:
$l={\int }_0^L\frac{\lambda gy}{AY}dy$

Integrating $y$ yields:
$l=\frac{\lambda g}{AY}{\left[\frac{y^2}{2}\right]}_0^L=\frac{\lambda g{L}^2}{2AY}$

To express this in terms of Young's modulus $Y$, we rearrange the equation:
$Y=\frac{\lambda g{L}^2}{2Al}$

Thus, the Young's modulus of the rod is $\frac{\lambda g{L}^2}{2Al}$.