A thin $1 \, m$ long rod has a radius of $5 \, mm$. A force of $50 \pi \times 10^{3} \, N$ is applied at one…

Given, $l=1 \mathrm{m}$, $r=5\times 10^{-3} \mathrm{m}$ and $F=50\text{π}\times 10^3 \text{N}$.

Least count in length is $=0.01 \mathrm{mm}=0.01\times {10}^{-3} \mathrm{m}$
Young's modulus is given by $Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}}$
For, $∆l\to \mathit{Min}$, $Y\to \mathit{Max}$
$\Rightarrow Y\frac{\Delta l}{l}=\frac{F}{A}$
$Y=\frac{50\text{π}\times 10^3}{\text{π}\times {\left(5\times 10^{-3}\right)}^2}\times \frac{l}{\Delta l}$
$Y=\frac{50\times 10^3}{25\times 10^{-6}}\times \frac{1}{\Delta l}\Rightarrow Y=\frac{2\times 10^9}{\Delta l}$
Here $\Delta l$ is unknown so we can not find the $Y$

The change in young’s modulus ratio is $\frac{∆Y}{Y}=2\frac{∆r}{r}+\frac{∆l}{l}$.

As seen from above, the minimum contribution to change in young’s modulus is given by the uncertainty in the length and the maximum contributing to the uncertainty in young’s modulus is given by uncertainty in radius.

Hence, correct option is $A$.