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

Solution :

Given :

Length of rod,

$$L = 1\text{ m}$$

Area of cross-section,

$$A = 3 \times 10^{-6}\text{ m}^2$$

Temperature fall,

$$\Delta T = 210^\circ\text{C} - 160^\circ\text{C}$$

$$= 50^\circ\text{C}$$

Coefficient of linear expansion,

$$\alpha = 2 \times 10^{-5}\text{ K}^{-1}$$

Young’s modulus,

$$Y = 2 \times 10^{11}\text{ N m}^{-2}$$

After cooling, rod contracts by :

$$\Delta L = \alpha L \Delta T$$

$$= 2 \times 10^{-5} \times 1 \times 50$$

$$= 10^{-3}\text{ m}$$

To regain original length, extension produced by attached mass must be :

$$\Delta L = 10^{-3}\text{ m}$$

Using Young’s modulus relation :

$$Y = \frac{\text{Stress}}{\text{Strain}}$$

$$Y = \frac{F/A}{\Delta L/L}$$

$$F = \frac{YA\Delta L}{L}$$

Substituting values :

$$F = \frac{(2 \times 10^{11})(3 \times 10^{-6})(10^{-3})}{1}$$

$$= 6 \times 10^2\text{ N}$$

Since,

$$F = Mg$$

$$M = \frac{600}{10}$$

$$= 60\text{ kg}$$

Final Answer :

$$60$$