Verified Solution • 2025–26 Session moderate

Q8.13: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.13 Solution: What is the density of water at a depth where pressure is 80.0…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often invert the volume‑change relation incorrectly, writing $ ho' =\rho(1-\Delta V/V)$ instead of the correct $ ho' =\rho/(1-\Delta V/V)\approx\rho(1+\Delta V/V)$. Remember that compression reduces volume, so density must increase; use the reciprocal of the volume factor, not a simple subtraction.

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03×103 kg m−31.03 \times 10^3 \text{ kg m}^{-3}?

Step-by-Step Verified Solution

**Answer:** ρ=1.034×103 kg/m3\rho = 1.034 \times 10^3 \text{ kg/m}^3 **Step-by-Step Explanation:** ΔP=80×1.013×105=8.10×106\Delta P = 80 \times 1.013 \times 10^5 = 8.10 \times 10^6 Pa. B=2.2×109B = 2.2 \times 10^9 Pa.
ΔVV=ΔPB=8.10×1062.2×109=3.68×10−3\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{8.10 \times 10^6}{2.2 \times 10^9} = 3.68 \times 10^{-3}
Since mass is conserved, ρ′=ρ/(1−ΔV/V)≈ρ(1+ΔV/V)\rho' = \rho/(1 - \Delta V/V) \approx \rho(1 + \Delta V/V).
ρ′=1.03×103(1+3.68×10−3)=1.034×103 kg/m3\rho' = 1.03 \times 10^3 (1 + 3.68 \times 10^{-3}) = 1.034 \times 10^3 \text{ kg/m}^3
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All Questions in this Chapter

Q8.1 Q8.2 Q8.3 Q8.4 Q8.5 Q8.6 Q8.7 Q8.8 Q8.9 Q8.10 Q8.11 Q8.12 Q8.13 Q8.14 Q8.15 Q8.16

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