A thin rod having a length of $1 \mathrm{~m}$ and area of cross-section $3 \times 10^{-6} \mathrm{~m}^{2}$ is suspended vertically from one end. The rod is cooled from $210^{\circ} \mathrm{C}$ to $160^{\circ} \mathrm{C}$. After cooling, a mass $\mathrm{M}$ is attached at the lower end of the rod such that the length of rod again becomes $1 \mathrm{~m}$. Young's modulus and coefficient of linear expansion of the rod are $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{-5} \mathrm{~K}^{-1}$, respectively. The value of $\mathrm{M}$ is __________ $\mathrm{kg}$.

(Take $\mathrm{g=10~m~s^{-2}}$)

When the rod is cooled from 210°C to 160°C, it will contract in length due to thermal contraction. The change in length of the rod is given by: <br/><br/>ΔL = L$\alpha$ΔT <br/><br/>where L is the original length of the rod, α is the coefficient of linear expansion, and ΔT is the change in temperature. <br/><br/>When a mass M is attached to the lower end of the rod, it will stretch due to the weight of the mass. The elongation of the rod is given by: <br/><br/>$\Delta L = {{MgL} \over {AY}}$ <br/><br/>where M is the mass, g is the acceleration due to gravity, A is the cross-sectional area of the rod, Y is the Young's modulus of the rod, and L is the original length of the rod. <br/><br/>$\therefore$ $L\alpha \Delta T = {{MgL} \over {AY}}$ <br/><br/>$\Rightarrow$ $\alpha \Delta T = {{Mg} \over {AY}}$ <br/><br/>$\Rightarrow$ $Mg = AY\alpha \Delta T$ <br/><br/>$\Rightarrow$ $\mathrm{M} \times 10=2 \times 10^{11} \times 3 \times 10^{-6} \times 2 \times 10^{-5} \times 50 $ <br/><br/>$\Rightarrow$ $\mathrm{M}=60 \mathrm{~kg}$