A thin 1m long rod has a radius of 5mm . A force of 50 π × 10 ^ 3 N...

Given, $l = 1m$, $r = 5 \times 1{0}^{- 3}m$ and $F = 50\text{\pi} \times 1{0}^{3}\text{N}$.

Least count in length is $= 0 . 01mm = 0.01 \times {10}^{- 3}m$
Young's modulus is given by $Y = \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}}$
For, $∆ l \to Min$, $Y \to Max$
$⇒ Y\dfrac{\Delta l}{l} = \dfrac{F}{A}$
$Y = \dfrac{50\text{\pi} \times 1{0}^{3}}{\text{\pi} \times {\left(5 \times 1{0}^{- 3}\right)}^{2}} \times \dfrac{l}{\Delta l}$
$Y = \dfrac{50 \times 1{0}^{3}}{25 \times 1{0}^{- 6}} \times \dfrac{1}{\Delta l} ⇒ Y = \dfrac{2 \times 1{0}^{9}}{\Delta l}$
Here $\Delta l$ is unknown so we can not find the $Y$

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

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 $A$.