Let $\alpha$ be the coefficient of linear expansion of iron and $\Delta T$ be the increase in temperature. If the original diameter is $D_0$, the new diameter is approximately $D = D_0(1 + \alpha \Delta T)$. The percentage increase in diameter is approximately proportional to $\alpha \Delta T$. If the original surface area is $A_0$, the new surface area is approximately $A = A_0(1 + 2\alpha \Delta T)$. The percentage increase in surface area is approximately proportional to $2\alpha \Delta T$. If the original volume is $V_0$, the new volume is approximately $V = V_0(1 + 3\alpha \Delta T)$. The percentage increase in volume is approximately proportional to $3\alpha \Delta T$. Density is mass per unit volume. Mass remains constant. As volume increases, density decreases. The question asks for percentage increase, so density is not the answer.
An iron ball is heated. The percentage increase will be largest in
Asked in WBPSC CS Prelims 2023 · Topic: Heat and Thermodynamics (Thermal Expansion (Solids, Liquids, Gases))
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