9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 liter, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Initial volume, V1 = 100 lit = 100 ×10-3 m3Final volume, V2 = 100.5 lit = 100.5 ×10-3 m3Increase in volume, ΔV = V2-V1 = 0.5 ×10-3 m3Increase in pressure, ΔP = 100 atm = 100 × 1.013 ×105PaBulk modulus, k = V1ΔPΔV = 100×10-3×100×1.013×1050.5×10-3 Pa = 2.026 ×109 PaBulk modulus of air = 1.0 ×105Pa(Bulk modulus of water / Bulk modulus of air) = (2.026 ×109)/ 1.0 ×105 = 2.026 ×104This higher ratio is attributed to the higher compressibility of air than water.

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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Solids

9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 liter, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

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Vishal Baghel | Contributor-Level 10

5 months ago

Initial volume, $V_{1}$ = 100 lit = 100 $\times 10^{- 3}$ $m^{3}$

Final volume, $V_{2}$ = 100.5 lit = 100.5 $\times 10^{- 3}$ $m^{3}$

Increase in volume, ΔV = $V_{2} - V_{1}$ = 0.5 $\times 10^{- 3}$ $m^{3}$

Increase in pressure, ΔP = 100 atm = 100 $\times$ 1.013 $\times 10^{5} P a$

Bulk modulus, k = $V_{1} \frac{Δ P}{Δ V}$ = $\frac{100 \times 10^{- 3} \times 100 \times 1.013 \times 10^{5}}{0.5 \times 10^{- 3}}$ Pa = 2.026 $\times 10^{9}$ Pa

Bulk modulus of air = 1.0 $\times 10^{5} P a$

(Bulk modulus of water / Bulk modulus of air) = (2.026 $\times 10^{9}) /$ 1.0 $\times 10^{5}$ = 2.026 $\times 10^{4}$

This higher ratio is attributed to the higher compressibility of air than water.

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