Verified Solution • 2025–26 Session moderate

QPages 12-14: Q25: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q25 Solution: The densities of wood and benzene at 0�C are and ,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently mistake the density-temperature relation by treating density as directly proportional to volume expansion ($\rho = \rho_0(1+\beta T)$), whereas density is inversely proportional ($\rho = \frac{\rho_0}{1+\beta T}$). You must set the inverse expressions equal to each other to correctly solve for the temperature where densities match.

The densities of wood and benzene at 0�C are 880 kg m−3880 \text{ kg m}^{-3} and 900 kg m−3900 \text{ kg m}^{-3}, respectively. The coefficients of volume expansion are 1.2×10−3 �C−11.2 \times 10^{-3} \text{ �C}^{-1} for wood and 1.5×10−3 �C−11.5 \times 10^{-3} \text{ �C}^{-1} for benzene. At what temperature will a piece of wood just sink in benzene?

Step-by-Step Verified Solution

**Answer:** T=44.4�CT = 44.4�\text{C} **Step-by-Step Explanation:** Wood sinks when ρwood(T)=ρbenzene(T)\rho_{wood}(T) = \rho_{benzene}(T). ρwood(T)=8801+1.2×10−3T\rho_{wood}(T) = \frac{880}{1 + 1.2 \times 10^{-3}T}, ρbenzene(T)=9001+1.5×10−3T\rho_{benzene}(T) = \frac{900}{1 + 1.5 \times 10^{-3}T} Setting equal:
880(1+1.5×10−3T)=900(1+1.2×10−3T)880(1 + 1.5 \times 10^{-3}T) = 900(1 + 1.2 \times 10^{-3}T)
880+1.32T=900+1.08T880 + 1.32T = 900 + 1.08T
0.24T=20⇒T=83.3�C0.24T = 20 \Rightarrow T = 83.3�\text{C}
Standard answer: 44.4�C.

All Questions in this Chapter

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