Physics Lab
NEET/Unit 7: Properties of Solids and Liquids/Young's, Bulk and Shear Moduli

Young's, Bulk and Shear Moduli

The three elastic moduli measure a solid's resistance to longitudinal stretching, uniform compression and shape distortion. NEET routinely tests their definitions, units and the relations among YY, KK and η\eta through Poisson's ratio σ\sigma.

Concept

For each kind of deformation, the modulus is the ratio of stress to strain within the elastic limit.

  • Young's modulus YY: longitudinal stress per unit longitudinal strain.
  • Bulk modulus KK: pressure per unit fractional volume change.
  • Shear (rigidity) modulus η\eta: tangential stress per unit shear angle.

Liquids and gases possess only a bulk modulus; they cannot sustain shear or longitudinal stress in static equilibrium. Compressibility is the reciprocal of KK.

Formula Derivation

The defining equations are

Y=FLA ΔL,K=−ΔPΔV/V,η=F/Aϕ.Y = \frac{F L}{A\,\Delta L}, \quad K = -\frac{\Delta P}{\Delta V/V}, \quad \eta = \frac{F/A}{\phi}.

Poisson's ratio is

σ=−lateral strainlongitudinal strain,0≤σ≤0.5.\sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}}, \quad 0 \le \sigma \le 0.5.

The three moduli are related by

Y=2η(1+σ)=3K(1−2σ),9Y=1K+3η.Y = 2\eta(1 + \sigma) = 3K(1 - 2\sigma), \quad \frac{9}{Y} = \frac{1}{K} + \frac{3}{\eta}.

For an isothermal ideal gas Kiso=PK_{\text{iso}} = P, while for an adiabatic process Kadi=γPK_{\text{adi}} = \gamma P.

NEET-style Worked Example

A solid copper cube of side 10 cm10\ \text{cm} is subjected to a uniform pressure of 107 Pa10^{7}\ \text{Pa}. If K=1.4×1011 PaK = 1.4 \times 10^{11}\ \text{Pa}, find the fractional change in volume.

ΔVV=−ΔPK=−1071.4×1011≈−7.14×10−5.\frac{\Delta V}{V} = -\frac{\Delta P}{K} = -\frac{10^{7}}{1.4 \times 10^{11}} \approx -7.14 \times 10^{-5}.

So ΔV≈−7.14×10−5×10−3 m3=−7.14×10−8 m3\Delta V \approx -7.14 \times 10^{-5} \times 10^{-3}\ \text{m}^{3} = -7.14 \times 10^{-8}\ \text{m}^{3}.

Common Confusions

  • YY, KK and η\eta all have units of pressure; only their physical meaning differs.
  • The negative sign in K=−ΔP/(ΔV/V)K = -\Delta P/(\Delta V/V) makes KK positive: increasing pressure shrinks volume.
  • Gases have Kadi=γKisoK_{\text{adi}} = \gamma K_{\text{iso}}, but the formula K=PK = P applies only to ideal gases at fixed temperature.
  • Liquids resist compression strongly: Kwater≈2.2×109 PaK_{\text{water}} \approx 2.2 \times 10^{9}\ \text{Pa} is far greater than KK for air.

Key Takeaways

  • YY for stretching, KK for compression, η\eta for shape.
  • Liquids: only KK. Gases: KK depends on the process.
  • Y=3K(1−2σ)=2η(1+σ)Y = 3K(1 - 2\sigma) = 2\eta(1 + \sigma) link the moduli through σ\sigma.
  • Compressibility =1/K= 1/K.
  • Kadi=γPK_{\text{adi}} = \gamma P for an ideal gas in adiabatic compression.

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