JEE (Advanced) · Physics, Class 11
Properties of Bulk Matter: JEE Advanced PYQ, 2007–2026
Properties of Bulk Matter (Physics, Class 11) has been asked 73 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.8 questions a paper, 10% of all Physics questions. It is the 3rd most-asked of 19 Physics chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 2in 2 papers |
| 2008 | 3in 2 papers |
| 2009 | 3in 2 papers |
| 2010 | 7in 2 papers |
| 2011 | 4in 2 papers |
| 2012 | 2in 2 papers |
| 2013 | 4in 2 papers |
| 2014 | 5in 2 papers |
| 2015 | 4in 2 papers |
| 2016 | 4in 2 papers |
| 2017 | 2in 2 papers |
| 2018 | 4in 2 papers |
| 2019 | 4in 2 papers |
| 2020 | 7in 2 papers |
| 2021 | 4in 2 papers |
| 2022 | 2in 2 papers |
| 2023 | 4in 2 papers |
| 2024 | 4in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 3in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Fluid mechanics22
- Heat transfer17
- Surface tension15
- Elasticity7
- Thermal expansion and calorimetry7
- Viscosity5
Try it
10 Properties of Bulk Matter questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Properties of Bulk Matter
A rod is stretched along its length. Its longitudinal strain is 5 × 10⁻⁴ and the lateral strain (contraction in width per unit width) is 1.5 × 10⁻⁴. What is the Poisson's ratio of the material?
Answer: (A)
Poisson's ratio is defined as the ratio of lateral strain to longitudinal strain, both measured as fractional changes. It has no units. Taking the magnitudes, σ = 1.5 × 10⁻⁴ / 5 × 10⁻⁴ = 0.3. The sign of the lateral strain is negative when the longitudinal strain is positive, but the Poisson's ratio is conventionally quoted as the positive magnitude.
Q2 · Physics · Properties of Bulk Matter
Two stars A and B have the same radius and both radiate as perfect black bodies. The wavelength at which star A emits most strongly is 400 nm, and for star B it is 800 nm. What is the ratio (total power radiated by A) : (total power radiated by B)?
Answer: (D)
Wien's displacement law states λ_m T = constant. So T_A/T_B = λ_B/λ_A = 800/400 = 2. Stefan–Boltzmann law gives power P = σAT⁴, and the radii are equal, so the areas are equal. P_A/P_B = (T_A/T_B)⁴ = 2⁴ = 16.
Q3 · Physics · Properties of Bulk Matter
A small spherical air bubble of radius 0.1 mm is held at a depth of 20 cm below the free surface of water in an open container. Take surface tension of water T = 0.07 N/m, density ρ = 1000 kg/m³ and g = 10 m/s². By how much does the pressure of the air inside the bubble exceed the atmospheric pressure?
Answer: (C)
The liquid just outside the bubble is at pressure P0 + ρgh. An air bubble in a liquid has one curved surface, so the pressure inside exceeds the outside liquid pressure by 2T/r. Hence P_inside − P0 = ρgh + 2T/r. Here ρgh = 1000 × 10 × 0.2 = 2000 Pa and 2T/r = 2 × 0.07/(1 × 10⁻⁴) = 1400 Pa. The total is 3400 Pa.
Q4 · Physics · Properties of Bulk Matter
Two solid spheres of the same material and surface emissivity have radii r and 2r. Both are at the same temperature T, well above that of the surroundings, and cool only by radiation. What is the ratio (initial rate of fall of temperature of the smaller sphere) : (initial rate of fall of temperature of the larger sphere)?
Answer: (A)
Power radiated is P = σeAT⁴, and the rate of temperature fall is dT/dt = P/(mc) = σeAT⁴/(mc). For spheres of the same material, A/m = 4πr² / ((4/3)πr³ρ) = 3/(ρr), which is proportional to 1/r. At the same T, the rate of fall is therefore inversely proportional to the radius. The ratio for r : 2r is 2 : 1.
Q5 · Physics · Properties of Bulk Matter
A liquid is filled in the gap between two large horizontal parallel plates. The lower plate is fixed and the upper plate moves with constant speed v. The gap holds two layers that do not mix. The upper layer is of thickness d and viscosity η, and it is in contact with the moving plate. The lower layer is of thickness 2d and viscosity 3η, and it is in contact with the fixed plate. The flow is steady and laminar, and the velocity varies linearly across each layer. What is the speed of the liquid at the interface between the layers?
Answer: (A)
In steady flow the shear stress must be the same in both layers, otherwise the interface would accelerate. Stress is η × (velocity change)/(thickness). Let u be the interface speed. The upper layer gives η(v − u)/d. The lower layer gives 3η(u − 0)/(2d). Setting them equal: 2(v − u) = 3u, so 2v = 5u and u = 2v/5.
Q6 · Physics · Properties of Bulk Matter
A siphon with a narrow tube of uniform cross-section carries water from a large open tank. Its highest point is 0.8 m above the water surface in the tank, and its free outlet is 1.8 m below that surface. Treat the water as ideal and take g = 10 m/s². What is the speed of the water leaving the outlet?
Answer: (D)
Apply Bernoulli's equation along a streamline from the tank surface (speed ≈ 0, pressure = atmospheric) to the outlet (pressure = atmospheric). The pressure terms cancel, leaving ½ρv² = ρg × (vertical drop from surface to outlet). So v = √(2 × 10 × 1.8) = √36 = 6 m/s. The height of the highest point does not affect the exit speed, as long as the siphon keeps working. It only limits how high the bend can be, since the pressure there falls.
Q7 · Physics · Properties of Bulk Matter
A wide open tank stands on a table 0.5 m above the ground and contains water to a depth of 1.0 m. A small hole at a depth of 0.6 m below the free surface sends a horizontal jet that lands on the ground at a certain point. A second small hole is to be made in the side wall so that its jet lands at the same horizontal distance from the tank. At what depth below the free surface should this second hole be made? (Ignore air resistance; the second hole is distinct from the first.)
Answer: (A)
Efflux speed from a hole at depth h is v = √(2gh). The jet then falls through the height y of the hole above the ground in time t = √(2y/g), so the range is x = v·t = 2√(h·y). The free surface is S = 1.0 + 0.5 = 1.5 m above the ground, so a hole at depth h has y = S − h and x = 2√(h(S − h)). This expression is unchanged under h → S − h. The first hole has h = 0.6 m, so the second has h = 1.5 − 0.6 = 0.9 m. This is less than 1.0 m, so the hole lies in the water. Check: 0.6 × 0.9 = 0.54 for both holes.
Q8 · Physics · Properties of Bulk Matter
A copper rod of length 0.40 m and cross-sectional area 1.0 cm² is lagged along its sides. One end is kept in steam at 100 °C and the other end is in contact with melting ice at 0 °C. In steady state, what mass of ice melts at the cold end in 60 minutes? (Thermal conductivity of copper = 400 W m⁻¹ K⁻¹; latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹; all heat conducted along the rod goes into melting ice.)
Answer: (A)
In steady state the heat current along a lagged rod is H = KAΔT/L. Here H = 400 × (1.0 × 10⁻⁴) × 100 / 0.40 = 10 W. In 60 min = 3600 s the heat conducted is Q = 10 × 3600 = 3.6 × 10⁴ J. This melts m = Q/L = 3.6 × 10⁴ / 3.36 × 10⁵ ≈ 0.107 kg of ice. The common slips are leaving the time in minutes, converting cm² wrongly, and mixing cal and J.
Q9 · Physics · Properties of Bulk Matter
A block of ice with a steel ball of mass 78 g frozen inside it floats in water in a vertical cylindrical vessel of cross-section 100 cm². The block melts completely and the steel ball sinks to the bottom. By how much does the water level in the vessel change? (Density of water = 1.0 g cm⁻³; density of steel = 7.8 g cm⁻³. Ignore evaporation and any temperature effects on density.)
Answer: (A)
By Archimedes' principle a floating body displaces water of weight equal to its own weight. While the ice and steel float together, the displaced volume is (M+m)/ρw. After melting, the ice becomes water of volume M/ρw, which adds exactly the ice's share of the displaced volume. The steel ball sinks and displaces only m/ρs. The change in total volume is therefore m/ρs − m/ρw = 78(1/7.8 − 1) = 10 − 78 = −68 cm³. Dividing by the area, 68/100 = 0.68 cm, so the level falls by 0.68 cm.
Q10 · Physics · Properties of Bulk Matter
A simple pendulum clock keeps correct time at 20 °C. The pendulum rod has coefficient of linear expansion 2 × 10⁻⁵ K⁻¹. The clock is moved to a room at 30 °C. Treating the pendulum length as that of the rod, how does the clock behave in one day (86400 s)?
Answer: (A)
The period is T = 2π√(L/g), so ΔT/T = ½ΔL/L = ½αΔθ = ½ × 2 × 10⁻⁵ × 10 = 10⁻⁴. A longer period means fewer ticks, so the clock runs slow. Time lost per day = 10⁻⁴ × 86400 s = 8.64 s.
Free with an account
All 73 real JEE Advanced questions from Properties of Bulk Matter.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
Solve them freeQuestions
About Properties of Bulk Matter in JEE Advanced.
How many questions come from Properties of Bulk Matter in JEE Advanced?
Properties of Bulk Matter (Physics, Class 11) has been asked 73 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.8 questions a paper, 10% of all Physics questions. It is the 3rd most-asked of 19 Physics chapters. The two 2026 papers had 3 from it.
Which topics of Properties of Bulk Matter are asked most in JEE Advanced?
Fluid mechanics (22), Heat transfer (17), Surface tension (15), Elasticity (7), counted over the papers from 2007–2026.
Is Properties of Bulk Matter important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 3 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Properties of Bulk Matter JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Properties of Bulk Matter with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
Your JEE Advanced 2027 plan, chapter by chapter.
A week-wise plan that gives the heavy chapters their time. Free, from a non-profit trust.