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PhysicsMediumClass 11

Bulk Modulus & Modulus of Rigidity

Properties of Matter

14

JEE Qs

8%

Hard

45

min

Master the definitions, distinguishing characteristics of the deformation each modulus describes, and the mathematical relationships between all elastic moduli for problem-solving efficiency.

🧮 Key Formulas

Bulk Modulus (K): K = -P / (ΔV/V) = -V (dP/dV)
Modulus of Rigidity (G or η): G = τ / φ
Pressure (P): P = F_normal / A
Shear Stress (τ): τ = F_tangential / A
Volume Strain: ΔV/V
Shear Strain (φ): φ = x/h (where x is relative displacement, h is height)
Relationship between Y, K, σ: Y = 3K(1 - 2σ)
Relationship between Y, G, σ: Y = 2G(1 + σ)
Relationship between K, G, σ, Y: 9/Y = 3/G + 1/K

✅ Key Points for JEE

  • 1Bulk Modulus (K) quantifies resistance to volume change under hydrostatic pressure. It is relevant for all states of matter (solids, liquids, gases).
  • 2Modulus of Rigidity (G) quantifies resistance to shape change (shear deformation) under tangential stress. It is primarily relevant for solids; liquids and gases have zero Modulus of Rigidity (G=0) as they cannot sustain shear stress.
  • 3The negative sign in the Bulk Modulus formula indicates that an increase in pressure (positive P) leads to a decrease in volume (negative ΔV).
  • 4All elastic moduli (Young's, Bulk, Rigidity) have the same dimensions and SI units as pressure (Pascal or N/m²).
  • 5The interrelationships between Young's Modulus (Y), Bulk Modulus (K), Modulus of Rigidity (G), and Poisson's ratio (σ) are crucial and frequently tested.

⚠️ Common Mistakes

  • Confusing Bulk Modulus with Young's Modulus or Modulus of Rigidity, especially their respective types of deformation (volume vs. longitudinal vs. shear).
  • Incorrectly using or interpreting the negative sign in the Bulk Modulus formula.
  • Not recognizing that liquids and gases only possess Bulk Modulus, as they do not resist shear forces.
  • Failing to recall or correctly apply the interrelationships between Y, K, G, and σ to solve problems.

📝 Practice Questions

See all

Q47.Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each other. The radius of curvature of the common surface, in cm , is ______ . and

2025·NumericalMedium

Q27.A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity) (1) 2 Mg (2) Mg (3) 3 2 Mg (4) Mg2

2025·MCQMedium

Q41. A tube of length L is shown in the figure. The radius of cross section at the point (1) is 2 cm and at the point (2) is 1 cm , respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be (1) 4 m/s (2) 2 m/s (3) 6 m/s (4) 8 m/s

2025·MCQEasy

Q30.A massless spring gets elongated by amount x1 under a tension of 5 N . Its elongation is x2 under the tension of 7 N . For the elongation of (5x1 −2x2), the tension in the spring will be, (1) 39 N (2) 15 N (3) 11 N (4) 20 N

2025·MCQEasy

Q32.Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P1 . The reading of the pressure gauge falls to P2 when the valve is opened. The speed of water flowing in the pipe is proportional to (1) P1 −P2 (2) (P1 −P2)4 (3) (P1 −P2)2 (4) √P1 −P2

2025·MCQMedium

Q48.An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 J/m2 and density 103 kg/m3 . The difference between pressure inside the bubble and atmospheric pressure is _____ N/m2 . (Take g = 10 m/s2 )

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 9: Mechanical Properties of Solids
  • Class 11 Physics Ch 10: Mechanical Properties of Fluids (for pressure and compressibility concepts)