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PhysicsMediumClass 11

Elasticity — Stress, strain, Young's modulus

Properties of Matter

14

JEE Qs

8%

Hard

75

min

Thoroughly understand the stress-strain curve for ductile and brittle materials, as its interpretation is a frequent source of conceptual and graphical questions.

🧮 Key Formulas

Stress (σ) = F/A
Longitudinal Strain (ε) = ΔL/L
Young's Modulus (Y) = Stress/Strain = (F/A) / (ΔL/L) = FL / (AΔL)
Hooke's Law: Stress ∝ Strain (within elastic limit)
Elastic Potential Energy Density (u) = 0.5 * Stress * Strain = 0.5 * Y * (Strain)^2
Total Elastic Potential Energy (U) = 0.5 * Stress * Strain * Volume = 0.5 * Y * (Strain)^2 * Volume

✅ Key Points for JEE

  • 1Stress is an internal restoring force per unit area, while pressure is an external force per unit area. Their units (Pascal or N/m^2) are the same but their physical interpretations differ.
  • 2Hooke's Law (Stress ∝ Strain) is valid only within the elastic limit. Beyond this, materials exhibit plastic deformation or fracture.
  • 3Young's Modulus (Y) is a material property that quantifies its resistance to deformation under tensile or compressive stress; it is independent of the sample's dimensions.
  • 4The stress-strain curve provides critical information about a material's behavior, including its elastic limit, yield strength, ultimate tensile strength, and fracture point.
  • 5Elastic potential energy stored in a deformed body is analogous to the potential energy of a spring, often asked as energy density or total energy.

⚠️ Common Mistakes

  • Confusing stress with pressure, especially in terms of application and internal/external nature.
  • Incorrectly identifying original length (L) and change in length (ΔL), or using inconsistent units for area and length.
  • Applying Hooke's Law or the Young's Modulus formula outside the elastic limit or for non-uniform stress distributions.
  • Forgetting to consider the cross-sectional area where the force is applied, especially for wires or rods.

📝 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