A thin rod of negligible mass and area of cross-section 4 x ${10}^{-6}{m}^2,$   suspended vertically from one end, has a length of 0.5 m at ${100}^0$  C. The rod is cooled to  $0^0$ C but prevented from contracting by attaching a mass at the lower end. The value of this mass is………………kg (Given, coefficient of linear expansion is  ${10}^{-5}$ ${ }^0{C}^{-1},$    Young's modulus is Y =  ${10}^{11}N{m}^{-2}$  and  g =10 $m{s}^{-2}$ ).
 

The change in length due to decrease in temperature,

$$\begin{gathered} \Delta l_1=L\alpha \Delta \theta =\left(0.5\right)\left({10}^{-5}\right)(0-100) \\ \Delta l_1=-0.5\times {10}^{-3}m\mathrm{...........}(i \end{gathered}$$

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 

$\Delta l_2=\frac{\left(Mg\right)L}{AY}=\frac{\left(M\right)\left(10\right)\left(0.5\right)}{(4\times {10}^{-6})\left({10}^{11}\right)}$  $\Delta l_2=(1.25\times {10}^{-5})M\mathrm{......}\left(ii\right)$

$$\begin{gathered} \Delta l_1+\Delta l_2=0 \Rightarrow (1.25\times {10}^{-5})M=(0.5\times {10}^{-3}) (From equations (i) and (ii\left)\right) \\ \Rightarrow M=\frac{0.5\times {10}^{-3}}{1.25\times {10}^{-5}}Kg \Rightarrow M=40Kg. \end{gathered}$$