Question

A wire of length 0.5 m and area of cross-section $4 \times 10^{-6}$ m$^2$ at a temperature of 100°C is suspended vertically by fixing its upper end to the ceiling. The wire is then cooled to 0°C, but is prevented from contracting, by attaching a mass at the lower end. If the mass of the wire is negligible, then the value of the mass attached to the wire is [Young's modulus of material of the wire = $10^{11}$ N/m$^2$; coefficient of linear expansion of the material of the wire = $10^{-5}$ K$^{-1}$ and acceleration due to gravity = 10 m/s$^2$]

• A. 10 kg
• B. 20 kg
• C. 30 kg
• D. 40 kg

Hint:

Thermal expansion: $\Delta L = L\alpha\Delta T$. Young's modulus: $Y = \frac{FL}{A\Delta L}$.

Answer 1

The Correct Option is D

Change in length due to temperature change $\Delta L = L\alpha\Delta T$, where $L$ is the original length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature. $\Delta L = (0.5)(10^{-5})(100) = 5 \times 10^{-4}$ m. Since the wire is prevented from contracting, the tension in the wire due to the attached mass produces an equal and opposite strain. Young's modulus $Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}$. The force $F$ is equal to the weight of the attached mass, $mg$. $10^{11} = \frac{mg(0.5)}{(4 \times 10^{-6})(5 \times 10^{-4})}$ $m(10) = \frac{10^{11} \times 4 \times 10^{-6} \times 5 \times 10^{-4}}{0.5} = 40$ $m = 40$ kg.