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Bulk Modulus (K) measures a material’s resistance to uniform compression. It is the ratio of the change in pressure to the resulting fractional change in volume.
K = ΔP / (ΔV / V)
K = - (ΔP · V) / ΔV (sign convention: ΔV negative under compression so K > 0)
Units: Pascal (Pa) where 1 Pa = 1 N/m². Liquids tend to have large K (nearly incompressible) compared with gases.
Work done (elastic energy) per unit volume
For small linear elastic deformations the energy stored per unit volume (elastic energy density) equals half the product of stress and strain (area under stress–strain curve):
U = ½ × (stress) × (strain) = ½ × (σ) × (ε) = ½ × (F/A) × (ΔL/L)
Applications & notes
Speed of sound in medium: v = √(K / ρ).
Liquids have large K; gases have much smaller K (more compressible).
K is reported as positive magnitude — the negative sign in formula accounts for ΔV being negative under compression.
10 JAMB Worked Examples
Example 1
Q: Cube: V = 1.00×10⁻³ m³. ΔP = 2.00×10⁵ Pa → ΔV = -1.00×10⁻⁶ m³. Find K.
Solution:
ΔV / V = (-1.00×10⁻⁶) / (1.00×10⁻³) = -1.00×10⁻³
K = ΔP / (ΔV / V) = 2.00×10⁵ / (−1.00×10⁻³) ⇒ magnitude = 2.00×10⁸ Pa
Example 2
Q: V = 0.020 m³ decreases by 2.0×10⁻⁵ m³ under ΔP = 1.0×10⁷ Pa. Find K.
ΔV / V = 2.0×10⁻⁵ / 0.020 = 1.0×10⁻³
K = 1.0×10⁷ / 1.0×10⁻³ = 1.0×10¹⁰ Pa
Example 3
Q: Wire: L = 2.00 m, A = 1.00×10⁻⁶ m², F = 5.00 N, ΔL = 2.00×10⁻³ m. Find U (J/m³).
σ = F/A = 5.00 / 1.00×10⁻⁶ = 5.00×10⁶ Pa
ε = ΔL / L = 2.00×10⁻³ / 2.00 = 1.00×10⁻³
U = ½ σ ε = 0.5 × 5.00×10⁶ × 1.00×10⁻³ = 2.50×10³ J/m³
Example 4
Q: K = 2.20×10⁹ Pa, ΔP = 1.10×10⁶ Pa. Find ΔV/V.
ΔV/V = ΔP / K = 1.10×10⁶ / 2.20×10⁹ = 5.00×10⁻⁴
Example 5
Q: Sphere: V = 0.500 m³ reduces by 5.00×10⁻⁵ m³. If K = 1.00×10⁹ Pa, find ΔP (magnitude).
ΔV/V = (-5.00×10⁻⁵)/0.500 = -1.00×10⁻⁴
ΔP = K (ΔV/V) ⇒ magnitude = 1.00×10⁵ Pa
Example 6
Q: Steel: K = 1.60×10¹¹ Pa, ρ = 7800 kg/m³. Find speed of sound v.
v = √(K/ρ) = √(1.60×10¹¹ / 7800) ≈ 4.53×10³ m/s
Example 7
Q: If U = 1000 J/m³ and σ = 2.00×10⁶ Pa, find ε.
ε = 2U / σ = 2×1000 / 2.00×10⁶ = 1.00×10⁻³
Example 8
Q: Liquid: K = 2.20×10⁹ Pa, ρ = 1000 kg/m³. Compute v.
v = √(K/ρ) = √(2.20×10⁹ / 1000) ≈ 1.48×10³ m/s
Example 9
Q: Block: V = 0.100 m³ → 0.0999 m³ under ΔP = 5.00×10⁶ Pa. Find K.
ΔV = -1.00×10⁻⁴ ⇒ ΔV/V = -1.00×10⁻³
K = 5.00×10⁶ / 1.00×10⁻³ = 5.00×10⁹ Pa
Example 10
Q: Spring: k = 200 N/m, x = 0.10 m. Es = ½ k x² = 1 J. L = 0.50 m, A = 1.00×10⁻⁴ m². Find U.
Es = ½ k x² = 0.5 × 200 × (0.10)² = 1.00 J
V ≈ A L = 1.00×10⁻⁴ × 0.50 = 5.00×10⁻⁵ m³
U = Es / V = 1 / 5.00×10⁻⁵ = 2.00×10⁴ J/m³
JAMB CBT Quiz — Bulk Modulus & Energy Density
Press Start Quiz. There is a 5-second countdown, then 20 minutes for 30 questions. The quiz auto-submits when time runs out. Attempts are saved to your browser.
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