NEET Physics — Properties of Matter Complete Chapter Guide

Chapter Overview & Weightage

Properties of Matter covers elasticity, viscosity, and surface tension — three distinct topics united by the study of how materials behave under forces. For NEET, this chapter contributes 1-2 questions per paper, usually on stress-strain relations, surface tension, or viscosity.

Properties of Matter carries 3-4% weightage in NEET. You can expect 1-2 questions. The most tested subtopics are: stress-strain curves, Young’s modulus, surface tension and capillarity, and Stokes’ law.

Year	NEET Questions	Key Topics Tested
2024	2	Surface tension, stress-strain
2023	1	Elastic moduli
2022	2	Capillarity, viscosity
2021	1	Young’s modulus
2020	2	Stokes’ law, surface energy

Key Concepts You Must Know

Tier 1 (Always asked):

Stress, strain, and their types (tensile, compressive, shear)
Young’s modulus, bulk modulus, shear modulus
Stress-strain curve interpretation (proportionality limit, elastic limit, yield point, breaking point)
Surface tension — definition, units, relation with surface energy
Capillarity — rise in narrow tubes

Tier 2 (Frequently asked):

Viscosity — Newton’s law of viscosity, coefficient of viscosity
Stokes’ law and terminal velocity
Excess pressure inside a drop and a bubble
Poisson’s ratio

Important Formulas

\text{Stress} = \frac{F}{A} \quad (\text{Unit: Pa or N/m}^2)

\text{Strain} = \frac{\Delta L}{L} \quad (\text{dimensionless})

Young’s Modulus: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{FL}{A\Delta L}$

Bulk Modulus: $B = \frac{-\Delta P}{\Delta V/V}$

Shear Modulus (Rigidity): $G = \frac{F/A}{\theta}$ where $\theta$ is the shear angle.

T = \frac{F}{L} \quad (\text{Unit: N/m})

Excess pressure inside a liquid drop:

\Delta P = \frac{2T}{R}

Excess pressure inside a soap bubble (two surfaces):

\Delta P = \frac{4T}{R}

Capillary rise:

h = \frac{2T\cos\theta}{\rho g r}

where $T$ = surface tension, $\theta$ = contact angle, $\rho$ = density, $r$ = radius of tube.

Newton’s law of viscosity:

F = \eta A \frac{dv}{dy}

Stokes’ law (drag force on a sphere in viscous fluid):

F = 6\pi\eta rv

Terminal velocity:

v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}

where $r$ = radius of sphere, $\rho$ = density of sphere, $\sigma$ = density of fluid, $\eta$ = viscosity.

For capillarity: water in a glass tube has $\theta \approx 0°$ (meniscus concave, rises). Mercury in glass has $\theta \approx 137°$ (meniscus convex, depresses). The sign of $\cos\theta$ determines rise or depression.

Solved Previous Year Questions

PYQ 1 — NEET 2024

Problem: A soap bubble of radius 1 cm is blown further to a radius of 2 cm. If surface tension $T = 0.04$ N/m, find the work done.

Solution:

Work done = change in surface energy = $T \times \Delta A$

A soap bubble has two surfaces (inner and outer), so:

W = T \times 2 \times 4\pi(R_2^2 - R_1^2)

W = 0.04 \times 2 \times 4\pi\left((0.02)^2 - (0.01)^2\right)

W = 0.04 \times 8\pi \times (4 \times 10^{-4} - 1 \times 10^{-4})

W = 0.04 \times 8\pi \times 3 \times 10^{-4}

\boxed{W = 9.6\pi \times 10^{-5} \approx 3.02 \times 10^{-4} \text{ J}}

The most common error: forgetting the factor of 2 for soap bubbles. A soap bubble has two free surfaces (inner and outer), so the total surface area change is $2 \times 4\pi(R_2^2 - R_1^2)$ . A liquid drop has only one surface.

PYQ 2 — NEET 2023

Problem: A wire of length 2 m and cross-sectional area $1 \times 10^{-6}$ m $^2$ is stretched by a force of 200 N. If $Y = 2 \times 10^{11}$ Pa, find the extension.

Solution:

Using $Y = \frac{FL}{A\Delta L}$ :

\Delta L = \frac{FL}{AY} = \frac{200 \times 2}{1 \times 10^{-6} \times 2 \times 10^{11}}

\boxed{\Delta L = \frac{400}{2 \times 10^5} = 2 \times 10^{-3} \text{ m} = 2 \text{ mm}}

PYQ 3 — NEET 2022

Problem: A steel ball of radius 2 mm falls through glycerine. If $\rho_{steel} = 8000$ kg/m $^3$ , $\rho_{glycerine} = 1300$ kg/m $^3$ , and $\eta = 0.83$ Pa.s, find the terminal velocity.

Solution:

v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}

v_t = \frac{2 \times (2 \times 10^{-3})^2 \times (8000 - 1300) \times 10}{9 \times 0.83}

v_t = \frac{2 \times 4 \times 10^{-6} \times 6700 \times 10}{7.47}

\boxed{v_t = \frac{0.536}{7.47} \approx 0.072 \text{ m/s} \approx 7.2 \text{ cm/s}}

Difficulty Distribution

Difficulty	% of Questions	What to Expect
Easy	40%	Direct formula application — Young’s modulus, surface tension units
Medium	40%	Capillarity calculations, Stokes’ law, stress-strain curve interpretation
Hard	20%	Energy stored in a stretched wire, combining elastic moduli, excess pressure problems

Expert Strategy

Priority order: Surface tension and capillarity first (most frequently tested), then elasticity (stress-strain), then viscosity (Stokes’ law). This chapter rewards formula memorization and careful unit work.

Key technique: Most problems are plug-and-chug — identify the correct formula, substitute values carefully. The only conceptual traps are the factor of 2 for soap bubbles and the sign of capillary rise for mercury.

Stress-strain curve questions are increasingly common. Remember the key points: proportionality limit, elastic limit, yield point, ultimate stress (tensile strength), and fracture point. Know what each represents qualitatively.

Common Traps

Trap 1 — Mixing up drop and bubble formulas. Excess pressure in a liquid drop: $2T/R$ . In a soap bubble: $4T/R$ . In an air bubble inside a liquid: $2T/R$ . The factor of 4 appears only for soap bubbles (two free surfaces).

Trap 2 — Using wrong units in terminal velocity. If radius is given in mm, convert to metres before substituting into Stokes’ formula. A factor of $10^{-3}$ gets squared to $10^{-6}$ — missing this gives an answer off by a million.

Trap 3 — Elastic energy. Energy stored per unit volume of a stretched wire = $\frac{1}{2} \times \text{stress} \times \text{strain}$ . Total energy = $\frac{1}{2} \frac{F^2 L}{AY}$ . Students often forget the $\frac{1}{2}$ factor.

Trap 4 — Poisson’s ratio. It is the ratio of lateral strain to longitudinal strain. For most materials, it ranges from 0 to 0.5. The theoretical maximum is 0.5 (for an incompressible material). Students sometimes confuse it with Young’s modulus.