NEET 2027 Physics: Mechanical Properties of Solids and Fluids & Thermodynamics Revision
Mishika Gupta17 Oct, 2026
Mechanical Properties of Fluids and Solids, along with Thermodynamics, can be challenging due to the number of formulas, numerical calculations and concepts involved. Confusion about sign conventions, graphs, pressure changes and thermodynamic conditions can lead to errors when solving NEET Physics questions.
These NEET 2027 notes bring together important concepts, formulas, solved numerical examples and revision points from all three topics. You can use them to understand key relationships, practise calculations and revise essential concepts before the examination.
Mechanical Properties of Fluids
Pressure, buoyancy, viscosity and surface tension are key concepts in the study of fluids. Revising their definitions, formulas and applications can help you solve numerical problems and prepare for NEET Physics.
1. Capillary Rise and Contact Angle
The height of capillary rise or fall is calculated using: [\boxed{h=\frac{2T\cos\theta}{\rho gr}}]
|
Symbol |
Meaning |
|
(h) |
Height of capillary rise or fall |
|
(T) |
Surface tension of the liquid |
|
(\theta) |
Angle of contact |
|
(\rho) |
Density of the liquid |
|
(g) |
Acceleration due to gravity |
|
(r) |
Inner radius of the capillary tube |
Important points:
-
A liquid that rises in a glass capillary generally has an acute contact angle and a concave meniscus.
-
Capillary rise increases with surface tension and decreases with the radius of the tube.
-
If (\theta<90^\circ), then (\cos\theta>0), giving a positive height of rise.
-
If (\theta>90^\circ), then (\cos\theta<0), indicating capillary depression.
Example: Finding the contact angle
If tangents at opposite sides of a meniscus form an angle of (60^\circ) and the arrangement is symmetric, the angle between each tangent and the horizontal is: [\theta=\frac{60^\circ}{2}=30^\circ]
2. Surface Tension and Surface Energy
Surface tension is associated with the energy required to increase a liquid’s surface area.
|
Concept |
Formula or Explanation |
|
Change in surface energy |
(\Delta E=T\Delta A) |
|
Surface energy of a spherical drop |
(E=4\pi r^2T), relative to a chosen zero of surface energy |
|
Effect of breaking a drop |
The total surface area increases, so additional energy is required when (T) is constant. |
Numerical formula: When a spherical liquid drop of radius (r) divides into (n) identical smaller drops, the increase in surface energy is: [\boxed{\Delta E=4\pi r^2T(n^{1/3}-1)}]
Here, (r) is the radius of the original drop.
Applications of surface tension:
-
Soap reduces the surface tension of water, helping it spread across fabrics.
-
Small liquid drops tend to become spherical because this shape minimises surface area for a fixed volume.
-
Breaking one drop into smaller drops increases the total surface area.
3. Viscosity and the Falling Ball Method
For a small sphere falling through a viscous liquid, Stokes’ law gives the terminal velocity:[\boxed{v_t=\frac{2r^2g(\rho_s-\rho_l)}{9\eta}}]
|
Symbol |
Meaning |
|
(v_t) |
Terminal velocity |
|
(r) |
Radius of the sphere |
|
(\rho_s) |
Density of the sphere |
|
(\rho_l) |
Density of the liquid |
|
(\eta) |
Coefficient of viscosity |
|
(g) |
Acceleration due to gravity |
Key relationships:
-
(v_t\propto r^2): terminal velocity increases with the square of the sphere’s radius.
-
(v_t\propto 1/\eta): terminal velocity decreases as viscosity increases.
-
The sphere must have reached terminal velocity for this formula to apply.
-
For liquids, viscosity generally decreases as temperature rises. For gases, it generally increases with temperature.
Stokes’ law applies under suitable low-speed flow conditions around a small sphere. An initial push may affect the time taken to reach terminal velocity, but it does not change the liquid’s viscosity.
4. Bernoulli’s Theorem and the Continuity Equation
Bernoulli’s theorem for steady flow of an ideal, incompressible fluid along a streamline is: [\boxed{P+\rho gh+\frac12\rho v^2=\text{constant}}]
For a horizontal pipe, the height terms cancel: P_2+\frac12\rho v_2^2 ]
The continuity equation is: [\boxed{A_1v_1=A_2v_2}]
|
Concept |
Important Relationship |
Explanation |
|
Bernoulli’s theorem |
(P+\rho gh+\frac12\rho v^2=\text{constant}) |
Relates pressure, height and fluid speed. |
|
Horizontal flow |
(P+\frac12\rho v^2=\text{constant}) |
Pressure and speed are related when height remains constant. |
|
Continuity equation |
(A_1v_1=A_2v_2) |
For steady incompressible flow, a smaller cross-sectional area means a higher speed. |
|
Pressure difference |
(\displaystyle P_1-P_2=\frac12\rho(v_2^2-v_1^2)) |
Relates pressure and velocity at two points in a horizontal pipe. |
Remember:
-
Use the continuity equation to relate velocities at different sections of a pipe.
-
Apply Bernoulli’s theorem to connect pressure, speed and height.
-
Two streamlines cannot intersect because the fluid velocity at one point cannot have two different directions at the same instant.
5. Buoyancy Involving Multiple Liquids
For a body floating at the boundary between two liquids, the buoyant force equals the weight of the displaced liquids.
[\boxed{mg=\rho_1V_1g+\rho_2V_2g}]
|
Symbol |
Meaning |
|
(m) |
Mass of the body |
|
(\rho_1,\rho_2) |
Densities of the two liquids |
|
(V_1,V_2) |
Volumes of the body immersed in the respective liquids |
|
(g) |
Acceleration due to gravity |
Example: Ice floating between water and kerosene
Given relative densities:
-
Ice = (0.9)
-
Kerosene = (0.8)
-
Water = (1.0)
Let (V_k) be the volume of ice immersed in kerosene and (V_w) the volume immersed in water.
For equilibrium: [0.9(V_k+V_w)=0.8V_k+1.0V_w]
Rearranging: [0.1V_k=0.1V_w]
Therefore: [\boxed{V_k=1:1}]
This result assumes the ice is in equilibrium at the liquid interface and the given relative densities apply.
6. Hydraulic Lift
A hydraulic lift works on Pascal’s law: pressure applied to an enclosed fluid is transmitted throughout the fluid.
[\boxed{P=\frac{F}{A}]
Example: Calculating the pressure required to support a vehicle
A vehicle of mass (5000,\text{kg}) rests on a piston with an area of (250,\text{cm}^2). Take (g=10,\text{m/s}^2).
|
Step |
Calculation |
Result |
|
1. Calculate the weight |
(F=mg=5000\times10) |
(50{,}000,\text{N}) |
|
2. Convert the area |
(250,\text{cm}^2=250\times10^{-4},\text{m}^2) |
(0.025,\text{m}^2) |
|
3. Calculate pressure |
(\displaystyle P=\frac{50{,}000}{0.025}) |
(2\times10^6,\text{Pa}) |
The required pressure is (2\times10^6,\text{Pa}), assuming the piston supports the vehicle’s entire weight and other forces are negligible.
Mechanical Properties of Solids
1. Young’s Modulus
Young’s modulus measures a material’s resistance to longitudinal deformation within its elastic limit.
[\boxed{Y=\frac{\text{Longitudinal stress}}{\text{Longitudinal strain]
|
Property |
Explanation |
|
Depends on |
Material and temperature |
|
Does not directly depend on |
Length and radius of the wire |
|
SI unit |
Pascal ((\text{Pa})) |
|
Physical significance |
A higher Young’s modulus indicates greater resistance to longitudinal deformation under the same stress. |
NEET point: If both the radius and length of a wire are doubled, Young’s modulus remains unchanged, provided the material and temperature stay the same.
2. Stress–Strain Graphs
The slope of a stress–strain graph depends on the quantities plotted on its axes.
|
Graph |
Slope |
Interpretation |
|
Stress versus strain |
(Y) |
Young’s modulus |
|
Strain versus stress |
(\frac{1}{Y}) |
Reciprocal of Young’s modulus |
For a strain-versus-stress graph, a smaller slope indicates a larger Young’s modulus.
Common mistake: Do not identify the material with the highest Young’s modulus simply by selecting the steepest line. Check the axes first.
3. Bulk Modulus
Bulk modulus measures resistance to uniform compression. [\boxed{K=-V\frac{dP}{dV]
The negative sign accounts for the decrease in volume as pressure increases.
Example: If pressure varies with volume as (P=aV^{-3})
|
Step |
Calculation |
|
Given relation |
(P=aV^{-3}) |
|
Differentiate |
(\displaystyle \frac{dP}{dV}=-3aV^{-4}) |
|
Substitute into the bulk modulus formula |
(\displaystyle K=-V(-3aV^{-4})) |
|
Simplify |
(K=3aV^{-3}) |
|
Since (P=aV^{-3}) |
(\boxed{K=3P}) |
Therefore, the bulk modulus is three times the pressure for the stated pressure–volume relationship.
Thermodynamics
Heat, work and temperature are closely related in thermodynamics, but their relationships change with the conditions of a process. Revising key laws, entropy, engine efficiency and energy conversion formulas can help you understand these concepts and tackle NEET Physics questions.
|
Topic |
Formula / Explanation |
Important Points |
|
1. Entropy Change During Heating |
For reversible heat transfer: (dS=\frac{\delta Q_{\mathrm{rev}}}{T}) For constant specific heat capacity: (\delta Q=mc,dT) Entropy change: (\displaystyle \Delta S=mc\ln\frac{T_2}{T_1}) |
Entropy measures energy dispersal and accessible microscopic arrangements. Use kelvin for both temperatures. The formula assumes constant specific heat capacity over the temperature range. Entropy and specific heat capacity are different physical quantities. |
|
2. First Law of Thermodynamics |
(\boxed{Q=\Delta U+W}) |
Here, (W) represents work done by the system. |
|
3. Isothermal Process |
For an ideal gas: (\Delta U=0), so (Q=W). For reversible expansion or compression: (\displaystyle \boxed{W=nRT\ln\frac{V_2}{V_1}}) |
Temperature remains constant, so the internal energy of an ideal gas does not change. Expansion gives positive work, while compression gives negative work. |
|
4. Adiabatic Process |
(\boxed{Q=0,\quad \Delta U=-W}) |
No heat enters or leaves the system. For an ideal gas, the internal energy change is related to its temperature change. |
|
5. Carnot Engine Efficiency |
(\displaystyle \boxed{\eta=1-\frac{T_c}{T_h}}) |
(T_h) is the hot-reservoir temperature and (T_c) is the cold-reservoir temperature, both in kelvin. Efficiency increases when (T_c) decreases while (T_h) remains constant. A real engine cannot exceed Carnot efficiency between the same temperatures. |
|
6. Intensive and Extensive Properties |
Extensive: Mass, volume, internal energy. Intensive: Pressure, temperature, density. |
Extensive properties depend on the amount of matter, while intensive properties are independent of the system size for a uniform system. |
|
7. Energy Conversion in an Insulated Vessel |
If all lost macroscopic kinetic energy becomes the gas's internal energy: (\displaystyle \frac12 Mv^2=nC_V\Delta T) For an ideal gas with (f) degrees of freedom: (\displaystyle C_V=\frac{fR}{2}) (\displaystyle \gamma=1+\frac{2}{f}) |
The equation assumes no energy loss to other forms and that the mass and gas quantity are defined consistently. |
|
8. Solved Example: Temperature Rise for (\gamma=1.4) |
Step 1: (\displaystyle 1.4=1+\frac{2}{f}) Step 2: (\displaystyle f=\frac{2}{1.4-1}=5) Step 3: (\displaystyle C_V=\frac{5R}{2}) Step 4: (\displaystyle \frac12 Mv^2=n\frac{5R}{2}\Delta T) Final result: (\displaystyle \boxed{\Delta T=\frac{Mv^2}{5nR}}) |
(M) is the mass whose kinetic energy is converted, (v) is its initial speed, (n) is the number of moles, and (R) is the universal gas constant. |
Revision Tips for NEET 2027
Use these revision tips to strengthen your understanding of key concepts, revise important formulas and improve your problem-solving accuracy for NEET 2027 Physics.
-
Mechanical Properties of Fluids: Revise capillary rise, surface tension, viscosity, Bernoulli’s theorem, buoyancy and Pascal’s law.
-
Mechanical Properties of Solids: Focus on Young’s modulus, bulk modulus and stress–strain graphs.
-
Thermodynamics: Practise questions on the first law, isothermal and adiabatic processes, entropy and Carnot engine efficiency.
-
Numerical Practice: Check units, sign conventions and the assumptions behind each formula before solving questions.
-
Formula Revision: Review important equations regularly and practise applying them to NEET-level problems.
Revising the key formulas and concepts of Thermodynamics can help you solve questions related to entropy, heat engines, energy changes and thermodynamic processes. Pay attention to sign conventions, temperature units and the assumptions used in each formula. Regular practice of numerical problems will help you apply these concepts accurately in NEET Physics.
Related Resources
Mechanical Properties of Fluids and Solids, and Thermodynamics FAQs
Why does heat equal work in an isothermal process?
Does Young’s modulus change when the dimensions of a wire are doubled?
What is the difference between intensive and extensive properties?
What is the efficiency formula for a Carnot engine?
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