Kumar Physics Classes
NEET PHYSICS TUTOR DOUBT 39
Physics practice paper for Gravitation, Mechanical Properties of Solids, Mechanical Properties of Fluids, Thermal Properties and Thermodynamics. Attempt every question like a real NEET paper, submit your answer, check the official solution, and track your score with NEET marking.
Kumar Physics Classes
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics increasingly rewards students who can connect concepts, formulas and numerical logic under exam pressure. Memorising a formula is not enough; students must know where it applies, what assumptions are hidden in the question, and how units, signs and graphs affect the final answer. A strong Physics foundation helps aspirants avoid careless mistakes and improves confidence across the full paper.
Important Message for NEET 2027 ,2028,2029,2030 and Future Aspirants
Future NEET aspirants should begin Physics preparation early and revise regularly. Chapters such as Gravitation, Solids, Fluids, Thermal Properties and Thermodynamics build the habit of analytical thinking. If these topics are practised step by step, students can improve calculation speed, formula selection and conceptual clarity long before the final examination.
Important Practice for Gravitation, SOLID fluid thermo
This practice set is designed to help students revise high-value concepts from Gravity, Solids, Fluids and Thermo in one focused worksheet. Attempt the questions first without looking at the solution, then use the answer and official solution buttons to correct your approach.
Physics Tutor Support for NEET Aspirants
If you are searching for a Physics Tutor for NEET and still facing difficulty in concepts, doubt-solving, numericals, or formula application, you may contact Kumar Sir. Kumar Sir explains Gravity, Solids, Fluids, Thermal Properties and other Physics topics in a very clear, step-by-step and exam-oriented way, helping students build confidence before tests.
Important Formula Revision for NEET Physics: Gravitation, Solids, Fluids and Thermodynamics
Before starting the question paper, revise these compact formulas carefully. Every formula is highlighted so that students can quickly recall the correct relation during numerical practice.
Gravitation
- Universal law: F = Gm₁m₂/r²
- Gravitational field: g = GM/r²
- Potential: V = -GM/r
- Potential energy: U = -GMm/r
- Escape velocity: vₑ = √(2GM/R)
- Orbital velocity: v₀ = √(GM/r)
- Satellite period: T = 2π√(r³/GM)
- Height variation: gₕ = g(1 - 2h/R), h << R
- Depth variation: g_d = g(1 - d/R)
Solids
- Stress: Stress = F/A
- Strain: Strain = ΔL/L
- Young's modulus: Y = Stress/Strain
- Bulk modulus: B = -VΔP/ΔV
- Compressibility: K = 1/B
- Shear modulus: η = Shear Stress/Shear Strain
- Elastic energy density: U = 1/2 × Stress × Strain
Fluids
- Liquid pressure: P = ρgh
- Buoyant force: F_B = ρVg
- Continuity: A₁v₁ = A₂v₂
- Bernoulli: P + 1/2ρv² + ρgh = constant
- Surface tension: T = F/l
- Liquid drop: ΔP = 2T/R
- Soap bubble: ΔP = 4T/R
- Terminal velocity: v_t = 2r²(ρ - σ)g/9η
- Poiseuille: Q = πPr⁴/8ηl
Thermal & Thermodynamics
- Heat: Q = mcΔT
- Latent heat: Q = mL
- Linear expansion: ΔL = αLΔT
- Area expansion: ΔA = βAΔT, β = 2α
- Volume expansion: ΔV = γVΔT, γ = 3α
- Conduction: H = KAΔT/L
- Cooling: dT/dt ∝ (T - T₀)
- Stefan: P = σAeT⁴
- Wien: λₘT = constant
- First law: ΔQ = ΔU + ΔW
- Gas work: W = PΔV
- Isothermal: PV = constant
- Adiabatic: PVγ = constant
- Mayer: Cₚ - Cᵥ = R
- Carnot efficiency: η = 1 - T₂/T₁
Students preparing for NEET must revise Gravitation, Solids, Fluids and Thermodynamics very carefully because these chapters often involve calculation, formula selection and conceptual confusion. Many students know the formulas but make mistakes while applying them in numerical questions. This question sheet has been designed to help students practise these topics in a focused and exam-oriented way. Every future NEET aspirant should attempt these questions seriously, check the solution carefully, and identify weak areas before the final exam.
If you are searching for a Physics Tutor for NEET and still facing difficulty in concepts, doubt-solving, numericals, or formula application, you may contact Kumar Sir. Kumar Sir explains Gravity, Solids, Fluids, Thermal Properties and other Physics topics in a very clear, step-by-step and exam-oriented way.
Kumar Physics Classes
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: kumarphysicsclasses.com
Question Index
Not attempted: cream | Selected: gold | Correct: green | Wrong: redQuestion 1+4 / -1
The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centres are at a distance d apart. The minimum speed with which a body of mass m should be projected from a point midway between the two centres so as to escape to infinity will be
Official PDF Solution: Using conservation of mechanical energy at the midpoint, the initial kinetic energy must balance the gravitational potential energy due to both bodies. This gives v = 2√[G(M₁ + M₂)/d].
Question 2+4 / -1
A satellite is revolving around a planet of mass M in an elliptical orbit of semi major axis a. The orbital speed of the satellite when it is at a distance 2a/3 from the planet will be. Consider the planet is at one focus of the orbit.
Official PDF Solution: From the vis-viva relation, v² = GM(2/r - 1/a). For r = 2a/3, v² = GM(3/a - 1/a) = 2GM/a, so v = √(2GM/a).
Question 3+4 / -1
Choose the incorrect statement among the following.
Official PDF Solution: Gravitational force is always attractive in nature. Therefore the statement that it becomes repulsive at large distances is incorrect.
Question 4+4 / -1
In a planetary motion, the areal velocity of a position vector of planet depends on angular velocity ω and distance of the planet from the sun r. The correct relation for areal velocity will be
Official PDF Solution: Areal velocity is dA/dt = L/2m. Since angular momentum L = mωr², dA/dt ∝ ωr².
Question 5+4 / -1
Two bodies of masses M1 and M2 are separated by a certain distance. If F₁₂ is the force on M1 due to M2 and F₂₁ is the force on M2 due to M1, then which relation is correct?
Official PDF Solution: The forces are equal in magnitude and opposite in direction: F₁₂ = -F₂₁. Hence F₁₂ + F₂₁ = 0 and |F₁₂| = |F₂₁|.
Question 6+4 / -1
The value of acceleration due to gravity on the surface of earth is x. At a depth d from the surface of earth, its value is y. If R is the radius of earth, then value of d will be
Official PDF Solution: For depth d, y = x(1 - d/R). Rearranging gives d = (1 - y/x)R.
Question 7+4 / -1
An infinite number of particles, each of mass 2 kg, are placed on the positive x-axis at distances 1 m, 4 m, 16 m, 64 m, ... respectively from the origin. The net gravitational potential at the origin due to the system of particles will be
Official PDF Solution: V = -GΣ(m/r) = -2G(1 + 1/4 + 1/16 + ...). The infinite geometric sum is 4/3, so V = -8G/3 J/kg.
Question 8+4 / -1
Let ω be the angular velocity of earth's rotation about its axis. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles, where h << R. The height h is, where g is acceleration due to gravity at poles.
Official PDF Solution: Equating apparent gravity at the equator and gravity at height h gives g(1 - 2h/R) = g - ω²R. Hence h = ω²R²/2g.
Question 9+4 / -1
A particle of mass M is placed at the centre of a uniform spherical shell of equal mass and radius a. The gravitational potential at point P at a distance a/3 from the centre will be
Official PDF Solution: Potential due to the central mass at r = a/3 is -3GM/a. Potential inside the spherical shell is -GM/a. Net potential is -4GM/a.
Question 10+4 / -1
A uniform spherical shell gradually shrinks maintaining its shape. The gravitational field at its centre will
Official PDF Solution: The gravitational field at every point inside a uniform spherical shell is zero. It remains zero even while the shell shrinks uniformly.
Question 11+4 / -1
If a planet of mass m is revolving around the sun in a circular orbit of radius 2r with time period T, then mass of the sun is
Official PDF Solution: For circular motion, T = 2π√(R³/GM). With R = 2r, M = 4π²(2r)³/GT² = 32π²r³/GT².
Question 12+4 / -1
The magnitude of gravitational potential energy of a body at a distance r from the centre of the earth is V. Its weight at a distance 3r from the centre of the earth is, where Re < r.
Official PDF Solution: Since V = GMm/r, the force at 3r is GMm/(3r)² = GMm/9r² = V/9r.
Question 13+4 / -1
N mole of a monoatomic gas undergoes a reversible triangular cycle ABCA, as shown in the figure. The work done by the gas in the cycle ABCA will be
Official PDF Solution: Work done in a cyclic P-V process equals the area enclosed by the cycle. The triangular area is P₀V₀/2.
Question 14+4 / -1
Heat is supplied at constant pressure to a polyatomic gas with γ = 4/3. The fraction of heat which goes to work done by the gas is
Official PDF Solution: For an isobaric process, W/Q = R/Cp = 1 - 1/γ. With γ = 4/3, W/Q = 1/4.
Question 15+4 / -1
Which of the following pressure-temperature (P-T) curve best represents the adiabatic process for an ideal gas?
Official PDF Solution: For an adiabatic process, P1-γTγ = constant. Since γ > 1, the P-T curve rises non-linearly as shown in option 1.
Question 16+4 / -1
The efficiency of a Carnot engine at a particular source and sink temperature is 2/3. If the sink temperature is reduced by 100°C, the engine efficiency becomes 3/4. The temperature of the source will be
Official PDF Solution: Using η = 1 - T₂/T₁ for both cases, 1 - T₂/T₁ = 2/3 and 1 - (T₂ - 100)/T₁ = 3/4. Solving gives T₁ = 1200 K.
Question 17+4 / -1
Which of the following does not depend on the path of the thermodynamic process?
Official PDF Solution: Internal energy is a state function. It depends only on initial and final states, not on the path.
Question 18+4 / -1
An ideal gas goes from A to B via two processes I and II as shown in the pressure-volume graph. If W₁ and W₂ are the work done by the gas in processes I and II respectively, then
Official PDF Solution: In both paths from A to B, the volume increases. Work done by the gas is positive when volume increases, so W₁ > 0 and W₂ > 0.
Question 19+4 / -1
Pressure-temperature (P-T) graph of an ideal diatomic gas is as shown in the figure. The molar heat capacity of the gas in the process will be, where R is universal gas constant.
Official PDF Solution: From the given process relation P ∝ 1/T and PV = nRT, the molar heat capacity for the process is obtained as C = 9R/2.
Question 20+4 / -1
Pressure-Volume (P-V) diagram of a thermodynamic process A → B → C is given below. Choose the correct Temperature-Volume (T-V) diagram.
Official PDF Solution: A → B is isochoric heating, so temperature increases at constant volume. B → C is isobaric expansion, so temperature increases with volume. This matches option 1.
Question 21+4 / -1
The heat supplied in the process 1 → 2 → 3 for two mole of monoatomic gas will be
Official PDF Solution: Heat is supplied in 1 → 2 and 2 → 3. For the isochoric part Q₁₂ = 3P₀V₀/2 and for the isobaric part Q₂₃ = 10P₀V₀. Total heat = 23P₀V₀/2.
Question 22+4 / -1
The work done by 2 moles of polyatomic gas (γ = 4/3), initially at room temperature, to increase its volume eight times during adiabatic process will be. Take R = 2 cal mol⁻¹ K⁻¹ and room temperature 27°C.
Official PDF Solution: For adiabatic expansion, TVγ-1 is constant. When V becomes 8V, T becomes 150 K. Using W = nRΔT/(1 - γ), W = 1800 cal.
Question 23+4 / -1
The coefficient of performance of a refrigerator is 4. If the temperature inside freezer is -23°C, then temperature of surrounding to which it rejects heat is
Official PDF Solution: COP = T₂/(T₁ - T₂). With T₂ = 250 K and COP = 4, T₁ = 312.5 K = 39.5°C.
Question 24+4 / -1
In the given Pressure-Volume (P-V) diagram, for the given mass of an ideal gas, the relation between temperatures T₁ and T₂ is
Official PDF Solution: For an ideal gas, PV = nRT. The isotherm marked T₁ lies at a higher temperature than T₂, so T₁ > T₂.
Question 25+4 / -1
If ΔU and ΔW represent the change in internal energy and work done by the system respectively, in thermodynamic process, then which of the following is incorrect?
Official PDF Solution: In an isobaric process, heat supplied is Q = ΔU + ΔW. Therefore ΔU = ΔW is not generally true.
Question 26+4 / -1
If Q, ΔU and W denote heat added, change in internal energy and work done in a cyclic process, then choose the correct option.
Official PDF Solution: In a cyclic process, initial and final states are the same, so ΔU = 0. From the first law, Q = W.
Question 27+4 / -1
A sample of gas is compressed from volume V₁ to V₂. The amount of work done on the gas is minimum when the compression is
Official PDF Solution: Work is represented by the area under the P-V curve. For the given comparison, the area is minimum for the isobaric path.
Question 28+4 / -1
The pressure of the gas filled in the bulb of a constant volume gas thermometer is 60 cm and 80 cm of mercury column at 0°C and 100°C respectively. If its bulb is immersed in a liquid at 80°C, then its pressure in mercury column will be
Official PDF Solution: For a constant volume gas thermometer, pressure varies linearly with temperature. (80/100) = (P - 60)/(80 - 60), so P = 76 cm.
Question 29+4 / -1
On the Kelvin scale temperature of the body increases by 40 K. The increase in temperature on the Fahrenheit scale will be
Official PDF Solution: A change of 40 K equals a change of 40°C. The Fahrenheit change is ΔF = (9/5)ΔC = 72°F.
Question 30+4 / -1
The mechanical equivalent of heat
Official PDF Solution: Mechanical equivalent of heat is the ratio of mechanical work to heat. Since both have dimensions of energy, the ratio is dimensionless.
Question 31+4 / -1
Some amount of steam at 100°C is passed into 40 g of water at 20°C and condensed into water. The final temperature of water becomes 40°C. The amount of steam converted into water will be. Take specific heat of water = 1 cal g⁻¹°C⁻¹ and latent heat of steam = 540 cal g⁻¹.
Official PDF Solution: Heat lost by steam = heat gained by water. m×540 + m×60 = 40×1×20, so 600m = 800 and m = 4/3 g.
Question 32+4 / -1
Two metal rods 1 and 2 of same length have same temperature difference between their ends. If the ratio of thermal conductivity of material of rods is 2:3 and ratio of their cross-sectional area is 2:1, then ratio of rate of heat conduction through them in steady state is
Official PDF Solution: Rate of heat conduction H = KAΔT/L. With same length and temperature difference, H₁:H₂ = (2×2):(3×1) = 4:3.
Question 33+4 / -1
The rate of energy radiated from a black body at 227°C is 20 W. If temperature of the black body is changed to 727°C, then its rate of energy radiated will be
Official PDF Solution: Radiated power is proportional to T⁴. Temperatures are 500 K and 1000 K, so power increases by 2⁴ = 16. New power = 320 W.
Question 34+4 / -1
The wavelength corresponding to maximum intensity of radiation from a black body at temperature 4000 K is 5000 Å. If temperature of body is decreased by 2000 K, then emitted wavelength corresponding to maximum intensity will be
Official PDF Solution: By Wien's displacement law, λT = constant. When temperature halves from 4000 K to 2000 K, wavelength doubles to 10000 Å.
Question 35+4 / -1
Instantaneous temperature difference between a body and surrounding, obeying Newton's law of cooling, is θ. Which option represents the variation of θ with time t, where θ₀ is initial temperature difference?
Official PDF Solution: Newton's law of cooling gives θ = θ₀e-bt. The temperature difference decays exponentially with time, matching option 2.
Question 36+4 / -1
A spherical air bubble is at the depth of 10 cm below the surface of water. If radius of the air bubble is 0.1 mm, then total pressure inside the air bubble will be nearly. Take surface tension of water 0.08 N/m, density of water 10³ kg/m³, atmospheric pressure 1.01 × 10⁵ Pa and g = 10 m/s².
Official PDF Solution: Pressure inside bubble = atmospheric pressure + hydrostatic pressure + excess pressure. P = 1.01×10⁵ + 0.1×10³×10 + 2×0.08/(0.1×10⁻³) ≈ 104 kPa.
Question 37+4 / -1
A liquid of density 2 × 10³ kg/m³ and coefficient of viscosity 3 × 10⁻² decapoise is flowing in a tube of radius 2 cm with speed 3/2 m/s. The flow of liquid in the tube is
Official PDF Solution: Using Reynolds number R = ρvD/η, the value comes out to 4000. Since R > 2000, the flow is turbulent.
Question 38+4 / -1
A liquid is coming out from two orifices of a fully filled tank kept on ground and falls on the same horizontal distance. If one orifice is x₁ above the mid-point of the tank and the other is x₂ below the mid-point of the tank, choose the correct relation between x₁ and x₂.
Official PDF Solution: For equal horizontal range, the two orifices must be symmetrically placed about the mid-point. Hence x₁ = x₂.
Question 39+4 / -1
A cube of edge length 20 cm is just balanced in two immiscible liquids A and B as shown in figure. If A and B have specific gravity 0.7 and 0.3 respectively, then mass of the cube will be
Official PDF Solution: In equilibrium, weight equals buoyant force. m = 0.2×0.2×0.06×700 + 0.2×0.2×0.14×300 = 3.36 kg.
Question 40+4 / -1
A wooden ball of density 400 kg/m³ immersed in water up to a depth d below the surface of water and then released. The acceleration of the ball inside the water will be, where g = 10 m/s².
Official PDF Solution: Net upward acceleration is a = (ρwater/ρball - 1)g = (1000/400 - 1)×10 = 15 m/s².
Question 41+4 / -1
Viscous drag on a spherical body is independent of
Official PDF Solution: By Stokes' law, viscous force F = 6πηrv. It depends on radius, velocity and viscosity, but not on the material of the body.
Question 42+4 / -1
In a hydraulic lift as shown in figure, mass of the car is m = 1000 kg, area of pistons A₁ = 20 cm², A₂ = 5 m². The minimum force F required to lift the car is, where g = 10 m/s².
Official PDF Solution: Using Pascal's law, F/A₁ = W/A₂. With W = 10000 N, A₁ = 20×10⁻⁴ m² and A₂ = 5 m², F = 4 N.
Question 43+4 / -1
A uniform rod of mass M and length L is placed on a smooth horizontal surface. If a horizontal force F is applied at one end as shown in the figure, then elongation produced in the rod will be. Y is Young's modulus and A is cross-sectional area.
Official PDF Solution: Tension varies linearly from zero at the free end to F at the pulled end. Integrating strain over the rod gives elongation = FL/2YA.
Question 44+4 / -1
If P, Q and R represent bulk modulus of elasticity for solid, liquid and gas respectively, then
Official PDF Solution: Bulk modulus of solids is greater than that of liquids, and liquids have greater bulk modulus than gases. Hence P > Q > R.
Question 45+4 / -1
A cube having edge length 6 cm. A tangential force of 12 kN, applied at its upper surface, displaces it by 0.2 mm relative to lower surface. The shear modulus of the material of cube will be
Official PDF Solution: Shear modulus η = FL/(AΔx). With F = 12×10³ N, L = 6×10⁻² m, A = 36×10⁻⁴ m² and Δx = 2×10⁻⁴ m, η = 1.0×10⁹ N/m².