Initial volume, V1 = 100.0l = 100.0 × 10 - 3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 -3 m3
Increase in volume, ΔV = V2 - V1 = 0.5 × 10 - 3 m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10 5 Pa
Bulk modulus = \(\frac{Δp }{ ΔV / V_1} =\frac{ Δp × V_1 }{ ΔV} \)
\(= \frac{100 × 1.013 × 10 ^5 × 100 × 10 ^{- 3 }}{ 0.5 × 106^{ - 3 }}\)
= 2.026 × 10 9 Pa
Bulk modulus of air = 1.0 × 10 5 Pa
\(∴\frac{ \text{Bulk modulus of water }}{ \text{Bulk modulus of air} }\)
\(= \frac{2.026 × 10 ^9 }{ 1.0 × 10^ 5 }\)
= 2.026 × 10 4
This ratio is very high because air is more compressible than water.
The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10–4 is ____ kJ/m3. Assume that material is elastic up to the linear strain of 5×10–4
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