Gravitation, Solids, Fluids and Thermal Properties Practice
This premium practice page brings together the Physics section questions Q1 to Q45 from the assessment PDF in a focused, mobile-friendly quiz format. Each question has clickable options, NEET marking, answer reveal, and the official solution matched with the correct question.
Why Strong Physics Preparation Is Now More Important Than Ever
NEET Physics rewards clarity, speed, and disciplined practice. Chapters such as gravitation, elasticity, fluids, thermal expansion, heat transfer, and surface tension appear simple at first glance, but they test whether a student can connect formulas with physical reasoning. A strong foundation now helps aspirants solve unfamiliar questions calmly, reduce silly mistakes, and protect marks when the paper becomes application-oriented.
Important Message for NEET 2027 and Future Aspirants
Future NEET aspirants should begin Physics preparation with consistency rather than pressure. Build concepts slowly, revise formulas weekly, and attempt mixed-topic quizzes after every chapter. The students who improve the most are not always those who study the longest; they are the ones who analyse mistakes, correct weak areas, and keep returning to standard concepts until they become natural.
Important Practice for Gravitation, SOLID and ThermAL PROPERTIES
This assessment is useful for strengthening high-yield Physics areas that often decide rank: gravitational potential and orbital motion, elastic stress and strain, capillarity, viscosity, Bernoulli's principle, thermal expansion, conduction, radiation, and calorimetry. Attempt every question seriously, check the solution only after making a choice, and revise the concept behind each incorrect answer.
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In a gravitational field, the gravitational potential is given by, V = -Kx (J/kg). The magnitude of gravitational field intensity at point (2, 0, 3) m is
Correct Answer: Option 3
Official PDF Solution
Hint: Use Ig = -∂V/∂x i - ∂V/∂y j - ∂V/∂z k.
Sol.:V = -Kx, hence Ig = -∂V/∂x i = Ki. Therefore |Ig| = K.
Q2
A uniform sphere of mass M and radius R is surrounded by a concentric spherical shell of same mass but radius 2R. A unit mass is kept at a distance x from centre where R < x < 2R in the region bounded by spheres as shown in the figure. The net gravitational potential at the location of unit mass is
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Correct Answer: Option 1
Official PDF Solution
Hint: Use the concept of gravitational potential due to a shell and a uniform solid sphere.
If an object is projected vertically upward from earth's surface with a speed that is one third of the escape speed from earth, then maximum height attained by it is [R is radius of earth]
Correct Answer: Option 3
Official PDF Solution
Hint: Use conservation of mechanical energy.
Sol.:Ki + Ui = Kf + Uf, with Ki = (1/2)m(ve/3)2, Ui = -GMm/R, Kf = 0, and Uf = -GMm/(R+h).
Solving gives R+h = 9R/8, hence h = R/8.
Q4
The linear speed of a planet in an elliptical orbit about the sun at positions A, B and C are vA, vB and vC respectively. If AC is the major axis and SB is perpendicular to AC at the position of the sun S as shown in figure, then
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Correct Answer: Option 1
Official PDF Solution
Hint: Use conservation of angular momentum.
Sol.:rC > rB > rA. Since mvr is constant, vC < vB < vA.
Q5
If the potential energy of a satellite in a circular orbit is -2 MJ, then the kinetic energy of the satellite is
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.: For a satellite in circular orbit, K.E. = -T.E. and T.E. = P.E./2.
Thus K.E. = -[-2]/2 = 1 MJ.
Q6
Time period of revolution of a satellite around a planet of radius R near its surface is T. Its period of revolution around another planet having radius 2R and same density (near the surface) will be
Correct Answer: Option 1
Official PDF Solution
Hint: Use T = 2πR/V0.
Sol.:V0 = √(GM/R), so T = 2πR3/2/√(GM). With M = (4/3)πR3ρ, T = √(3π/(ρG)).
For the same density, time period is independent of radius of planet.
Q7
The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 6R for same planet will be
A body cools down from 70°C to 60°C in 5 minutes in a particular ambient temperature. The time it will take to cool down from 60°C to 50°C in the same ambient temperature will be
Correct Answer: Option 3
Official PDF Solution
Hint: Use Newton's law of cooling.
Sol.: Rate of cooling is proportional to the excess of mean temperature over ambient temperature. For 70°C to 60°C, R1 ∝ [65 - T0]; for 60°C to 50°C, R2 ∝ [55 - T0]. Hence R1 > R2.
For the same drop in temperature, the time taken would be more for 60°C to 50°C.
Q9
A constant volume thermometer registers a pressure of 1.5 × 104 Pa at 273.16 K and a pressure of 2.05 × 104 Pa at the normal boiling point of water. The temperature at the normal boiling point is
Correct Answer: Option 1
Official PDF Solution
Hint: At constant volume, pressure is directly proportional to temperature.
Sol.:P ∝ T. Thus 1.5×104 / 2.05×104 = 273.16 / T, so T = (2.05/1.5)(273.16) = 373.3 K.
Q10
Which of the following graphs shows the most appropriate variation of volume of 1 kg of water with the increase in temperature?
Correct Answer: Option 2
Official PDF Solution
Hint: As the temperature is increased from 0°C to 4°C, the volume of water decreases.
Sol.: The volume of water is minimum at 4°C. Between 0°C to 4°C it decreases and after 4°C it increases.
Q11
The neck and bottom of a bottle are 2 cm and 10 cm in radius respectively. If the cork is pressed with a force of 12 N in the neck of bottle, then force exerted on the bottom of bottle is
Correct Answer: Option 3
Official PDF Solution
Hint: Use Pascal's law.
Sol.:12/[π(2)2] = F/[π(10)2], hence 12/4 = F/100 and F = 300 N.
Q12
Consider the following statements: Statement (A): When height of tube is less than liquid rise in capillary tube, the liquid does not overflow. Statement (B): Product of radius of meniscus and height of liquid in capillary tube always remains constant. In the light of above statements, choose the correct answer.
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.:h = 2S/(Rρg). The liquid does not overflow because hR = h'R' is constant.
Thus liquid rises to full length of tube and radius of curvature becomes larger.
Q13
A force of magnitude F is applied at an angle θ from horizontal on the free end of a cylindrical bar as shown. The shear stress developed at the given cross-section of area A will be
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Correct Answer: Option 4
Official PDF Solution
Hint: Shear stress = tangential force / area of cross-section.
Sol.: Tangential force is F sinθ. Therefore shear stress = F sinθ/A.
Q14
A steel wire 4 m in length is stretched through 2 mm. The cross-sectional area of wire is 2 mm2. If Young's modulus of steel is 2 × 1011 N m-2, then the stress produced in wire is
The ratio of tensile stress to longitudinal strain is termed as
Correct Answer: Option 2
Official PDF Solution
Hint & Sol.: Young's modulus = normal stress / longitudinal strain.
Q16
A wire of initial length L and radius r is stretched by a length l. Another wire of same material but with initial length 2L and radius 2r is stretched by a length 2l. The ratio of stress produced in both wires is
Correct Answer: Option 4
Official PDF Solution
Hint: Young's modulus = stress / strain.
Sol.: Strain produced in both wires is same. Material is same; hence stress produced will be same.
Q17
A vertical hanging bar of length l and mass m per unit length carries a load of mass M at the lower end, its upper end is clamped to a rigid support. The maximum tensile stress developed in the bar is (A → area of cross-section of bar)
Correct Answer: Option 4
Official PDF Solution
Hint: The tensile stress produced due to weight of bar would be non-uniform.
Sol.: Maximum stress is produced at the fixed end of the bar. σmax = Mg/A + (m×l×g)/A = (M + ml)g/A.
Q18
The excess pressure in a soap bubble is thrice of that in other one. The ratio of their volume is
Correct Answer: Option 1
Official PDF Solution
Hint: Excess pressure inside a soap bubble, ΔP = 4T/r.
Sol.:4T/r1 = 3×4T/r2, so r2/r1 = 3. Since volume = (4/3)πr3, ratio of volume is 1 : 27.
Q19
A square shaped wire frame of side length R is floating on the surface of liquid of surface tension S. The force required to just pull out the frame from the liquid is
Correct Answer: Option 3
Official PDF Solution
Hint & Sol.: Force required to just pull it out = 2 × S × 4R = 8SR.
Q20
The weight of a body in water is fifty percent of its weight in air. The density of the body is
A beaker containing a liquid of density ρ moves down with an acceleration a where a < g. The gauge pressure due to liquid at a depth h below free surface of liquid is
Correct Answer: Option 2
Official PDF Solution
Hint & Sol.: Effective acceleration geff = g - a when the beaker moves down.
Therefore pressure P = ρh(g - a).
Q22
If two soap bubbles of different radius are connected by a tube, then
Correct Answer: Option 3
Official PDF Solution
Hint & Sol.: Since pressure inside the bubble of smaller radius is more, air flows from the smaller radius bubble to the bigger radius bubble.
Q23
The velocity of the surface layer of water in river of depth 5 m is 10 m s-1. The shearing stress between the surface layer and the adjacent layer will be (Coefficient of viscosity of water, η = 10-3 kg m-1 s-1)
A capillary tube is dipped in a liquid as shown in figure. Let the pressure at points A, B and C be PA, PB and PC respectively, then
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Correct Answer: Option 4
Official PDF Solution
Hint & Sol.:PA = PC = P0 (atmospheric pressure), and P0 - PB = 2T/r.
Thus P0 > PB, so PA > PB and PC > PB.
Q25
In case of an ideal fluid, Bernoulli's theorem expresses the application of the principle of conservation of
Correct Answer: Option 2
Official PDF Solution
Hint & Sol.: Bernoulli's theorem is based on the principle of conservation of energy.
Q26
What will be the ratio of the acceleration due to gravity and gravitational constant if M is assumed to be the mass of earth and R to be its radius?
Correct Answer: Option 2
Official PDF Solution
Hint & Sol.:g = GM/R2. Therefore g/G = M/R2.
Q27
The mass of a planet is 1/5th of that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is
Correct Answer: Option 2
Official PDF Solution
Hint: Acceleration due to gravity on surface of earth, g = GM/R2.
Sol.: For the planet, g' = G(M/5)/(R/2)2 = (4/5)×9.8 = 7.84 m s-2.
Q28
If a liquid drop spreads on a glass plate then the angle of contact between liquid and glass can be
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.: A water drop spreads on a glass plate because the angle of contact is acute.
Q29
The energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of R from surface of earth is
E - GMm/R = -GMm/(2R) + (1/2)m v02, so E = GMm/R - GMm/(2R) + GMm/(4R) = 3GMm/(4R).
Q30
Which of the given processes can result in liberation of heat?
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.: The condensation of vapour into liquid can result in liberation of heat.
Q31
If two rods of length L and 2L having coefficient of linear expansion α and 2α respectively are connected so that total length becomes 3L, then the equivalent coefficient of linear expansion of composite rod will be
The rate of flow of heat through a metal bar of area of cross-section 1 m2 when temperature gradient is 1°C/m under steady state is equal to
Correct Answer: Option 2
Official PDF Solution
Hint: Use Fourier's law of heat conduction.
Sol.:ΔQ/Δt = K A (ΔT/Δx). If ΔT/Δx = 1°C/m and A = 1 m2, then ΔQ/Δt = K.
Q33
Two slabs are connected as shown having same length and same cross-sectional area. In steady state, the temperature of their interface will be
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Correct Answer: Option 4
Official PDF Solution
Hint: In steady state the rate of flow of heat current will be same through both the slabs.
Sol.:(T - 40)KA/l = (100 - T)2KA/l. Hence T - 40 = 200 - 2T, so 3T = 240 and T = 80°C.
Q34
Heat is being supplied at a constant rate to a sphere of ice and it melts completely in 100 s. The rise of temperature thereafter will be
Correct Answer: Option 1
Official PDF Solution
Hint: No temperature change takes place during melting.
Sol.: Let the constant rate be α. For melting, m×80/α = 100, so m = 100α/80. After melting, α = m×1×ΔT, hence ΔT = α/m = 0.8°C/s.
Q35
If the temperature of a black body is increased by 33.3% then the amount of radiation emitted by it will
Correct Answer: Option 1
Official PDF Solution
Hint: Rate of radiation is proportional to the fourth power of absolute temperature.
Sol.:R ∝ T4. If T' = T + T/3 = 4T/3, then R2/R1 = (4/3)4 = 3.16. Percentage increase = 216%.
Q36
The pressure that has to be applied to the ends of a steel rod of length 10 cm to keep its length constant when its temperature is raised by 50°C is (For steel Young's modulus is 2 × 1011 N m-2 and coefficient of thermal expansion is 1.1 × 10-5 K-1.)
Correct Answer: Option 2
Official PDF Solution
Hint: The sum of thermal strain is equal to longitudinal strain to keep length constant.
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in capillary tube is 5m. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is
Correct Answer: Option 3
Official PDF Solution
Hint: The force due to surface tension balances the weight of liquid column.
Sol.:mg = 2πrT cosθ, so mass of water is proportional to radius of capillary. When radius becomes 2r, mass becomes 10m.
Q38
Consider two statements given below out of which one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A): A thin stainless steel needle can lay floating on a still water surface. Reason (R): Thin stainless steel needle floats when buoyancy force balances its weight.
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.: Needle floats due to the force of surface tension.
Q39
Water enters a horizontal pipe of non-uniform cross-section with a velocity of 0.5 m/s and leaves the other end with a velocity of 0.7 m/s. If the pressure of water at first end is 103 N/m2, then the pressure at other end is
The liquid surfaces have a tendency to contract. This phenomenon may be attributed to
Correct Answer: Option 1
Official PDF Solution
Hint & Sol.: The liquid surfaces have a tendency to contract due to surface tension.
Q41
If acceleration due to gravity at a height h is g/4, then the value of h in terms of radius of earth Re is
Correct Answer: Option 4
Official PDF Solution
Hint: Acceleration due to gravity at height h is g' = g/(1 + h/R)2.
Sol.:g/4 = g/(1 + h/R)2, hence h = Re.
Q42
A simple pendulum has a time period T1 when on earth's surface and T2 when taken to height R/2 above the earth's surface, where R is radius of earth. The ratio T2/T1 is equal to
Correct Answer: Option 4
Official PDF Solution
Hint: Time period of simple pendulum T = 2π√(l/geff).
