Verified Solution • 2025–26 Session moderate

Q8.12: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.12 Solution: Compute the bulk modulus of water from the following data:…

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing $B=\Delta P/(\Delta V/V)$, both $\Delta V$ and $V$ must be expressed in the same SI unit (m$^3$); using litres for one and cubic metres for the other (or treating 0.5 L as 0.5 m$^3$) inflates the denominator by a factor of $10^3$ and gives a bulk modulus off by three orders of magnitude.

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 _ 10_ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Step-by-Step Verified Solution

**Answer:** Bwater=2.026×109B_{water} = 2.026 \times 10^9 Pa. Ratio Bwater/Bair≈104B_{water}/B_{air} \approx 10^4. **Step-by-Step Explanation:** ΔP=100×1.013×105=1.013×107\Delta P = 100 \times 1.013 \times 10^5 = 1.013 \times 10^7 Pa. ΔV=100.5−100.0=0.5\Delta V = 100.5 - 100.0 = 0.5 L =0.5×10−3= 0.5 \times 10^{-3} m_. V=100V = 100 L =0.1= 0.1 m_.
B=ΔPΔV/V=1.013×1070.5×10−3/0.1=1.013×1075×10−3=2.026×109 PaB = \frac{\Delta P}{\Delta V/V} = \frac{1.013 \times 10^7}{0.5 \times 10^{-3}/0.1} = \frac{1.013 \times 10^7}{5 \times 10^{-3}} = 2.026 \times 10^9 \text{ Pa}
For air, B≈105B \approx 10^5 Pa. The ratio is about 10410^4. Water is much less compressible than air because its molecules are closely packed [citation:16].

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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