A thin rod having a length of 1m and area of cross-section 3 × 10 ^...

Let $\Deltal$ be the decease in length of rod due to decease in temperature

Then, $\Deltal = l\alpha\DeltaT$, here, coefficient of linear expansion is $\alpha = 2 \times {10}^{- 5}{K}^{- 1}$ and $\DeltaT = ( 483 - 433 ) K$$= 50K$

Thus, $\Deltal = 1 \times 2 \times {10}^{- 5} \times 50 = {10}^{- 3}m$

Young's modulus is stated as $Y = \dfrac{\dfrac{F}{A}}{\dfrac{\Deltal}{l}}$, here, $F = Mg$.

So, $2 \times {10}^{11} = \dfrac{Mg \times 1}{3 \times {10}^{- 6} \times {10}^{- 3}}$

$⇒ Mg = 2 \times {10}^{11} \times 3 \times {10}^{- 9} = 6 \times {10}^{- 2}$

Thus, $M = 60kg$