A thin rod of negligible mass and area of cross-section 4 × 10 ^ -...

The change in length due to decrease in temperature,

$\Delta {\text{l}}_{1} = \text{L}\alpha \Delta \theta = \left(0 \cdot 5\right)\left({10}^{- 5}\right)\left(0 - 100\right)$

$\Delta {\text{l}}_{1} = - 0 \cdot 5 \times {10}^{- 3}\text{m}$ ... (i)

Negative sign implies that length is decreasing. Now let $M$ be the mass attached to the lower end. Then, change in length due to suspension of load is

\Delta {\text{l}}_{2} = \dfrac{\left(\text{Mg}\right)\text{L}}{\text{AY}} = \dfrac{\left(\text{M}\right)\left(10\right)\left(\text{0.5}\right)}{\left(4 \times {10}^{- 6}\right)\left({10}^{11}\right)}$<br /><br />$\Delta {\text{l}}_{2} = \left(\text{1.25} \times {10}^{- 5}\right)M$... (ii)<br /><br />$\Delta {\text{l}}_{1} + \Delta {\text{l}}_{2} = 0$<br /><br />$⇒ \left(\text{1.25} \times {10}^{- 5}\right)M = \left(\text{0.5} \times {10}^{- 3}\right)$(from equations (i) and (ii))<br /><br />$⇒ M = \dfrac{\text{0.5} \times {10}^{- 3}}{\text{1.25} \times {10}^{- 5}}kg$<br /><br />$⇒ M = 40kg