A thin rod having a length of 1 m and area of cross-section 3 × 10⁻⁶ m² is suspended vertically from one end. The rod is cooled from 210 to 160. 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 × 10¹¹ N m⁻² and 2 × 10⁻⁵ K⁻¹, respectively. The value of M is __________ kg. (Take g = 10 m s⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 60. Contraction on cooling: Δ L = LαΔ T = 1 × 2 × 10⁻⁵ × 50 = 10⁻³ m. The load must stretch it back by the same amount: Mg = (YAΔ L)/(L) Mg = 2 × 10¹¹ × 3 × 10⁻⁶ × 10⁻³ = 600 N ⇒ M = 60 kg So the answer is 60 kg.

A thin rod having a length of 1 m and area of cross-section 3 × 10⁻⁶… | JEE Main 2023 (31 Jan, Shift 1)

Q6JEE Main 2023 · 31 JAN · SHIFT 1Difficult

Thermal Stress & Breaking Stress

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

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 60

Contraction on cooling: $\Delta L = L\alpha\Delta T = 1 \times 2 \times 10^{-5} \times 50 = 10^{-3}\text{ m}$ .
The load must stretch it back by the same amount: $Mg = \dfrac{YA\Delta L}{L}$
$Mg = 2 \times 10^{11} \times 3 \times 10^{-6} \times 10^{-3} = 600\text{ N} \Rightarrow M = 60\text{ kg}$
So the answer is 60 kg.

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