Gravitation, Bulk Matter, Thermal Properties and Thermodynamics Practice Paper
Attempt the Physics section Q1 to Q45 with NEET marking, instant checking, official solution support and a clean final score summary. The page is designed for quick practice, revision and confidence building before a serious test attempt.
This practice page brings together important Physics questions from Gravitation, Properties of Bulk Matter, Thermal Properties of Matter and Thermodynamics. Use it like a real paper: revise once, attempt honestly, check only after selecting an option, and then study the official solution carefully.
Physics understanding also grows step by step as concepts become clearer. Every solved question, every corrected mistake, and every understood concept adds value to a student's preparation. The main purpose of these questions is to help students build real conceptual clarity, so that Physics does not remain a burden but becomes a subject they can understand, enjoy, and solve with confidence.
Why Strong Physics Preparation Is Now More Important Than Ever
Physics rewards students who can connect formulas with physical meaning. In present NEET preparation, simply remembering results is not enough; students must know when a formula applies, what each term represents and how to move from a concept to a calculation without panic. Regular practice with mixed questions trains accuracy, time management and exam temperament together.
For NEET 2027 and Future Batches
Important Message for NEET 2027 and Future Aspirants
Students preparing for NEET 2027 and later should start building clarity early. Do not wait for the syllabus to feel complete before solving meaningful questions. A steady routine of concept revision, formula recall, numerical practice and mistake analysis creates a strong base. When fundamentals are handled patiently, advanced questions become less frightening and scoring in Physics becomes much more realistic.
Concept Practice
Important Practice for Gravitation, Bulk Matter and Thermodynamics
These chapters contain many scoring ideas: fields and potentials, elasticity, fluids, heat transfer, calorimetry and thermodynamic processes. The best way to master them is to revise formulas briefly, solve questions in exam mode, and then compare your reasoning with the official solution. Each attempt should improve both speed and conceptual discipline.
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Kumar Physics Classes helps NEET aspirants strengthen concepts through focused explanation, doubt support and exam-oriented practice. Students get a clear path for understanding formulas, applying them in numerical questions and improving accuracy through regular correction. If you are preparing for NEET 2027 or any future medical entrance attempt, consistent Physics guidance can make your preparation more organised, less stressful and more result-focused. Use this paper sincerely, note your mistakes, revise the formulas and contact Kumar Sir when you need personal direction for your next step.
Important Formula Revision Before Starting the Paper
Before solving the paper, revise these high-yield formulas from Gravitation, Properties of Bulk Matter, Thermal Properties of Matter and Thermodynamics. Keep the ideas fresh, then attempt the questions with speed and accuracy.
Two particles of equal mass m go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
Correct Answer: Option 3
Official Solution: Centripetal force is provided by the gravitational force of attraction between the two particles: mv2/R = Gm2/(2R)2. Therefore v = 1/2 √(Gm/R).
Question 2
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The escape velocity for a planet is ve. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
Correct Answer: Option 1
Official Solution: From mechanical energy conservation, 0 = 1/2 mv2 - 3GMm/2R. Hence v = √(3GM/R) = √1.5 ve.
Question 3
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A particle is projected vertically upwards from the surface of the earth (radius Re) with a speed equal to one fourth of escape velocity. What is the maximum height attained by it from the surface of the earth?
Correct Answer: Option 2
Official Solution: Using conservation of mechanical energy, 1/2 mv2 = GMm/Re - GMm/R, where R is maximum distance from earth's centre. Also v = ve/4. Solving gives R = 16Re/15, so h = R - Re = Re/15.
Question 4
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Gravitational potential difference between a point on the surface of a planet and another point 10 m above is 4 J/kg. Considering gravitational field to be uniform, how much work is done in moving a mass of 2 kg from the surface to a point 5 m above the surface?
Correct Answer: Option 1
Official Solution: Gravitational field g = -ΔV/Δx = 4/10 J kg-1 m-1. Work done in moving 2 kg through 5 m is W = mgh = 2 × (4/10) × 5 = 4 J.
Question 5
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The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B, then
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Correct Answer: Option 4
Official Solution: From Kepler's law, areal velocity is constant. Therefore Area(SCD)/t1 = Area(SAB)/t2. Since Area(SCD) = 2 Area(SAB), t1 = 2t2.
Question 6
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Which of the following statements is true about acceleration due to gravity by Earth?
Correct Answer: Option 3
Official Solution: Variation of g with distance shows that at the centre of the earth the acceleration due to gravity is zero. Also, rotational correction is g' = g - ω2Rcos2λ.
Question 7
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An astronaut, inside an earth satellite experiences weightlessness because
Correct Answer: Option 1
Official Solution: The astronaut's acceleration is g, so he is in free fall. Therefore the floor of the satellite exerts no normal reaction on him and he experiences weightlessness.
Question 8
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Three identical point masses, each of mass 1 kg lie in the x-y plane at points (0, 0), (0, 0.2 m) and (0.2 m, 0). The gravitational force on the mass at the origin is
Correct Answer: Option 3
Official Solution: Each force has magnitude G(1)(1)/(0.2)2 = 6.67 × 10-11/0.04 = 1.67 × 10-9 N. Hence Fnet = Fî + Fĵ = 1.67 × 10-9(î + ĵ) N.
Question 9
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The gravitational field due to a mass distribution is E = K/x3 in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at the distance x is
Correct Answer: Option 4
Official Solution: Using ∫dV = -∫E dx, 0 - V = -∫x∞K/x3 dx. Therefore V = K/(2x2).
Question 10
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Two bodies of masses m and M are placed at distance d apart. The gravitational potential (V) at the position where the gravitational field due to them is zero is
Correct Answer: Option 4
Official Solution: For the neutral point, distances are in the ratio √m : √M. Adding potentials of both masses gives V = -(G/d)(√m + √M)2.
Question 11
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A satellite is seen after each 8 hours over equator at a place on the Earth when its sense of rotation is opposite to the Earth. The time interval after which it can be seen at the same place when the sense of rotation of Earth and satellite is same will be
Correct Answer: Option 3
Official Solution: For opposite directions the relative angular speed is ω1 + ω2 and the period is 8 h. With ω1 = 2π/24 and ω2 = π/6, for same direction the relative speed is ω2 - ω1, giving t = 24 h.
Question 12
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Two identical satellites are at the heights R and 7R from the Earth's surface. Then which of the following statement is incorrect (r = radius of the Earth)?
Correct Answer: Option 1
Official Solution: For circular orbit, K.E. = GMm/2r, P.E. = -GMm/r, and T.E. = -GMm/2r. With orbital radii 2R and 8R, ratios of K.E., P.E. magnitude and total energy magnitude are 4:1. Hence statement (1) is incorrect.
Question 13
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During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Density of blood = 1.06 × 103 kg m-3 and g = 9.8 m/s2].
Correct Answer: Option 1
Official Solution: Pressure P = hρg. Thus h = 2000/(1.06 × 103 × 9.8) = 0.192 m.
Question 14
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A tank full of water has a small hole at its bottom. Let t1 be the time taken to empty first one third of the tank, t2 be the time taken to empty second one third of the tank and t3 be the time taken to empty rest of the tank, then
Correct Answer: Option 3
Official Solution: As water height decreases, efflux speed and rate of flow decrease. Therefore successive equal volume portions take more time: t1 < t2 < t3.
Question 15
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A piece of cork starts from rest at the bottom of a lake and floats up. Its velocity v is plotted against time t. Which of the following best represents the resulting curve?
Correct Answer: Option 1
Official Solution: As the cork moves up, buoyant force remains constant. As speed increases, viscous retarding force increases in proportion to speed, so acceleration gradually decreases and velocity approaches a limiting value. This matches option (1).
Question 16
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A 2 m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radians. The shear strain developed will be
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
Correct Answer: Option 2
Official Solution: For same material and tension, Δl = FL/AY, so Δl ∝ L/r2. The wire with 100 cm length and 1 mm diameter gives the largest value.
Question 18
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An increase in pressure required to decrease the 200 litres volume of a liquid by 0.004% in container is (Bulk modulus of the liquid = 2100 MPa)
Correct Answer: Option 4
Official Solution: Bulk modulus B = ΔP/(-ΔV/V). Hence ΔP = 2100 × 106 × (0.004/100) = 84 kPa.
Question 19
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The radius of a soap bubble is r. The surface tension of soap solution is T. Keeping temperature constant, the radius of the soap bubble is doubled, the energy necessary for this will be
Correct Answer: Option 1
Official Solution: Initial surface energy of a soap bubble is 2T × 4πr2 = 8πr2T. Final energy at radius 2r is 32πr2T. Required energy is 24πr2T.
Question 20
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A soap bubble in vacuum has a radius of 3 cm and another soap bubble in vacuum has a radius of 4 cm. If the two bubbles coalesce under isothermal condition, then the radius of the new bubble is
Correct Answer: Option 3
Official Solution: For coalescing soap bubbles under isothermal condition, rnew2 = r12 + r22 = 32 + 42, so rnew = 5 cm.
Question 21
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In a capillary tube experiment, a vertical 30 cm long capillary tube is dipped in water. The water rises up to a height of 10 cm due to capillary action. If this experiment is conducted in a freely falling elevator, the length of the water column becomes
Correct Answer: Option 3
Official Solution: In a freely falling elevator, effective gravity is zero. Since h = 2Tcosθ/rdgeff, water rises to the maximum available height, equal to the capillary length: 30 cm.
Question 22
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A tank of height 5 m is full of water. There is a hole of cross sectional area 1 cm2 in its bottom. The initial volume of water that will come out from this hole per second is (g = 10 m/s2)
Correct Answer: Option 1
Official Solution: Velocity of efflux v = √(2gh) = √(2 × 10 × 5) = 10 m/s. Rate of flow Q = Av = 1 × 10-4 × 10 = 10-3 m3/s.
Question 23
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Two drops of equal radius are falling through air with a steady velocity of 5 cm/s. If the two drops coalesce, then its terminal velocity will be
Correct Answer: Option 1
Official Solution: When two equal drops coalesce, R = 21/3r. Since terminal velocity vT ∝ r2, v'T/vT = (R/r)2 = 22/3 = 41/3. Therefore v'T = 41/3 × 5 cm/s.
Question 24
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The total weight of a piece of wood is 6 kg. In the floating state in water its 1/3 part remains inside the water. On this floating wood, what maximum weight is to be put such that the whole of the piece of wood is to be downed in the water?
Correct Answer: Option 1
Official Solution: In floating condition, 6g = ρwg(V/3). When fully immersed with added mass m, (6 + m)g = ρwgV. Comparing gives 18 = 6 + m, so m = 12 kg.
Question 25
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An aeroplane of mass 3 × 104 kg and total wing area of 120 m2 is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings (in kilo pascal) is (g = 10 m/s2)
Correct Answer: Option 1
Official Solution: Force due to pressure difference is ΔP × A. In balanced level flight, mg = ΔP A. Thus ΔP = (3 × 104 × 10)/120 = 2.5 kPa.
Question 26
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Water stands upto a height h behind the vertical wall of a dam. What is the net horizontal force pushing the dam down by the stream, if width of the dam is σ? (ρ = density of water)
Correct Answer: Option 2
Official Solution: Average pressure on the dam wall is hρg/2. Area is hσ. Net horizontal force F = (hρg/2)(hσ) = h2ρgσ/2.
Question 27
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Water rises in a capillary upto a height h. If now this capillary is tilted by an angle of 45°, then the length of the water column in the capillary becomes
Correct Answer: Option 4
Official Solution: Vertical height remains h. If the tube is inclined at 45°, length in capillary l = h/cos45° = h√2.
Question 28
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In a 20 m deep lake, the bottom is at a constant temperature of 4°C. The air temperature is constant at -10°C. If the thermal conductivity of ice is 4 times that of water, neglecting the expansion of water on freezing, the maximum thickness of ice will be
Correct Answer: Option 2
Official Solution: In steady state the heat flow through water and ice is equal: KA(4 - 0)/(20 - x) = 4KA[0 - (-10)]/x. Solving gives x = 200/11 m.
Question 29
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The temperature of a body rises by 44°C when a certain amount of heat is given to it. The same heat when supplied to 22 g of ice at -8°C, raises its temperature by 16°C. The water equivalent of the body is [Given: swater = 1 cal/g°C, Lf = 80 cal/g, sice = 0.5 cal/g°C]
Correct Answer: Option 2
Official Solution: Heat supplied to ice: 22 × 0.5 × 8 + 22 × 80 + 22 × 1 × 16 = 88 + 1760 + 352 = 2200 cal. Heat capacity of body = 2200/44 = 50 cal/°C. Water equivalent = 50/1 = 50 g.
Question 30
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A refrigerator converts 100 g of water at 25°C into ice at -10°C in one hour and 50 minutes. The quantity of heat removed per minute is (specific heat of ice = 0.5 cal/g°C, latent heat of fusion = 80 cal/g)
Correct Answer: Option 2
Official Solution: Heat removed cooling water to 0°C: 100 × 1 × 25 = 2500 cal. Freezing heat: 100 × 80 = 8000 cal. Cooling ice to -10°C: 100 × 0.5 × 10 = 500 cal. Total 11000 cal in 110 min, so 100 cal/min.
Question 31
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540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is (Given latent heat of fusion of ice = 80 cal/g and specific heat capacity of water = 1 cal/g°C)
Correct Answer: Option 1
Official Solution: Heat required to melt ice: Q = mL = 540 × 80 = 43200 cal. Heat released by water cooling to 0°C: 540 × 1 × 80 = 43200 cal. It is just sufficient to melt all ice, so final temperature is 0°C.
Question 32
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A brass rod of cross-sectional area 1 cm2 and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young's modulus of elasticity of brass is 1 × 1011 N/m2 and g = 10 m/s2, then increase in the energy of the rod will be
Correct Answer: Option 2
Official Solution: Increase in elastic energy = F2L/(2AY) = (5 × 10)2 × 0.2 / [2 × 10-4 × 1011] = 2.5 × 10-5 J.
Question 33
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The unit of Stefan's constant is
Correct Answer: Option 4
Official Solution: From Stefan's law, P = σAT4. Therefore [σ] = W m-2K-4.
Question 34
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Which one of the following process depends on gravity?
Correct Answer: Option 3
Official Solution: Convection depends on buoyancy and gravity, while conduction and radiation do not require gravity.
Question 35
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The ratio of coefficient of thermal conductivity of two different materials is 5 : 3. If the thermal resistance of the two rods of these materials of same area is same, then the ratio of the length of these rods will be
Correct Answer: Option 1
Official Solution: Thermal resistance R = l/kA. Since R and A are same, l1/l2 = k1/k2 = 5/3.
Question 36
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On centigrade scale the temperature of a body increases by 30 degrees. The increase in temperature on Fahrenheit scale is
Correct Answer: Option 4
Official Solution: A temperature difference of 100°C equals 180°F. Therefore 30°C corresponds to (180/100) × 30 = 54°F.
Question 37
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A constant volume gas thermometer shows pressure reading of 50 cm and 90 cm of mercury at 0°C and 100°C respectively. When the pressure reading is 60 cm of mercury, the temperature is
The coefficient of volume expansion of a liquid is 49 × 10-5 K-1. Calculate the fractional change in its density when the temperature is raised by 30°C.
Correct Answer: Option 3
Official Solution: When temperature rises, V' = V(1 + γΔT), so ρ' = ρ/(1 + γΔT). Fractional change = 1 - 1/(1 + γΔT) = γΔT/(1 + γΔT). With γΔT = 49 × 10-5 × 30 = 0.0147, value is about 1.5 × 10-2.
Question 39
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A hammer of mass 1 kg having speed of 50 m/s, hit an iron nail of mass 200 g. If specific heat of iron is 0.105 cal/g°C and half the energy is converted into heat, the raise in temperature of nail is
Correct Answer: Option 1
Official Solution: Half the kinetic energy is converted into heat: (1/2)(1/2)Mv2 = JmcΔθ. Thus (1/4) × 1 × 502 = 4.2[200 × 0.105 × Δθ], giving Δθ = 7.1°C.
Question 40
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1 gram of ice at 0°C is mixed with 1 gram of steam at 100°C. At thermal equilibrium, the temperature of the mixture is
Correct Answer: Option 1
Official Solution: Heat released by steam on condensation is mLv = 540 cal. Heat required to melt ice is mLf = 80 cal. Sufficient heat remains to raise water to 100°C, so final temperature is 100°C.
Question 41
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Surface of the lake is at 2°C. Find the temperature of the bottom of the lake
Correct Answer: Option 3
Official Solution: The densest water layer is at the bottom. Water has maximum density at 4°C, so the bottom of the lake is at 4°C.
Question 42
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A metal plate 4 mm thick has a temperature difference of 32°C between its faces. It transmits 200 kcal/h through an area of 5 cm2. Thermal conductivity of the material is
Correct Answer: Option 1
Official Solution: Here Δx = 4 × 10-3 m, ΔT = 32°C, heat current ΔQ/Δt = 200 × 1000 × 4.2 / 3600 = 233.33 J/s, and A = 5 × 10-4 m2. From ΔQ/Δt = KA(ΔT/Δx), K = 58.33 W m-1°C-1.
Question 43
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Consider a compound slab consisting of two different material having equal thickness and thermal conductivities k and 3k respectively as shown in figure. The equivalent thermal conductivity of the slab is
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Correct Answer: Option 1
Official Solution: The two layers are in series: R = R1 + R2. Therefore 2l/(keffA) = l/(kA) + l/(3kA). Hence 2/keff = 4/3k, so keff = 3k/2.
Question 44
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Three rods of the same dimensions have thermal conductivities 3k, 2k and k. They are arranged as shown, with their ends at 100°C, 50°C and 0°C. The temperature of their junction is
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Correct Answer: Option 2
Official Solution: Let θ be the junction temperature. Net heat current at the junction is zero: 3k(100 - θ) + k(0 - θ) + 2k(50 - θ) = 0. Solving gives θ = 200/3°C.
Question 45
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A 10 kg body falls from a height of 10 m. If its entire potential energy is converted into heat and absorbed by 2 kg of water (specific heat of water = 4200 J kg-1°C-1, take g = 10 m/s2), the rise in temperature of water is
Correct Answer: Option 4
Official Solution: Official solution context: entire P.E. gets converted into heat. ΔP.E. = msΔθ. Therefore 10 × 10 × 10 = 2 × 4200 × Δθ, so Δθ = 0.12°C.
