Previous year questions: Mechanical Properties of Fluids

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the formulas on the first page of this chapter, and every answer is worked out in full so that the method behind the correct option is clear. The surface tension questions need the area change of a sphere, the viscosity question needs Stokes' law at terminal speed, and the buoyancy question needs only the displaced volume.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A liquid drop of diameter 2 mm breaks into 512 identical droplets. The change in surface energy is alpha x 10^-6 J. The surface tension of the liquid is 0.08 N m^-1. Find alpha.

Answer: The original radius is R = 1 mm. Volume is conserved, so 512 r^3 = R^3 and each droplet has radius r = R/8. The change in area is 512 x 4 pi r^2 - 4 pi R^2 = 4 pi R^2 (512/64 - 1) = 4 pi R^2 x 7. With R = 10^-3 m this is 4 pi x 10^-6 x 7 = 8.8 x 10^-5 m^2. Multiplying by the surface tension gives 0.08 x 8.8 x 10^-5 = 7.03 x 10^-6 J, so alpha is about 7. The option 7 is correct.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A soap bubble of diameter 2 cm is blown to a diameter of 6 cm. The surface tension of the soap solution is 0.03 N m^-1. The work done is alpha pi x 10^-4 J. Find alpha. Take pi = 3.14.

Answer: A soap bubble has two surfaces, so the area change is 2 x 4 pi (r2^2 - r1^2), with r1 = 1 cm and r2 = 3 cm. Then r2^2 - r1^2 = 9 x 10^-4 - 1 x 10^-4 = 8 x 10^-4 m^2, and the area change is 2 x 4 pi x 8 x 10^-4 = 64 pi x 10^-4 m^2. Work = T x area change = 0.03 x 64 pi x 10^-4 = 1.92 pi x 10^-4 J. Comparing with alpha pi x 10^-4 J gives alpha = 1.92.

Q3 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A 1 mm radius air bubble moves up through a liquid of density 2000 kg m^-3 with a steady speed of 0.5 cm s^-1. Take g = 10 m s^-2 and neglect the air density on the buoyancy side. Work out the viscosity of the liquid in poise.

Answer: At the terminal speed the viscous drag balances the buoyant force minus the tiny weight of the air. Using 6 pi eta r v = (4/3) pi r^3 sigma g, the coefficient is eta = 2 r^2 sigma g/(9 v). With r = 10^-3 m and v = 5 x 10^-3 m s^-1, eta = 2 x 10^-6 x 2000 x 10/(9 x 5 x 10^-3) = 0.888 Pa s. One poise is 0.1 Pa s, so eta is about 8.8 poise. The option 8.8 is correct.

Q4 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A soap bubble of radius 1 cm is blown to a radius of 2 cm. The surface tension of the soap solution is 3.5 x 10^-2 N m^-1. The work required is alpha x 10^-6 J. Find alpha. Take pi = 22/7.

Answer: The bubble has two surfaces. The area change is 2 x 4 pi (0.02^2 - 0.01^2) = 8 pi x (4 x 10^-4 - 1 x 10^-4) = 8 pi x 3 x 10^-4 = 24 pi x 10^-4 m^2. Work = 3.5 x 10^-2 x 24 x (22/7) x 10^-4 = 3.5 x 10^-2 x 75.4 x 10^-4 = 2.64 x 10^-4 J. Hence alpha = 264.

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Eight mercury drops, each of radius r, coalesce into one big drop. The surface tension of mercury is S. Find the surface energy released.

Answer: Volume is conserved, so the big drop has radius R with R^3 = 8r^3, giving R = 2r. The total surface area before coalescence is 8 x 4 pi r^2 = 32 pi r^2, and after it is 4 pi R^2 = 16 pi r^2. The area released is 16 pi r^2, so the energy released is 16 pi r^2 S. The option 16 pi r^2 S is correct.

Q6 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A wooden cube of side 10 cm floats in water with part of it submerged. The density of wood is 0.4 g cm^-3 and that of water is 1 g cm^-3. A metal coin is placed on the cube, and the submerged part increases by 3.87 cm. Find the mass of the coin in grams.

Answer: The extra immersed depth is 3.87 cm over a face area of 10 cm x 10 cm = 100 cm^2, so the extra displaced volume is 387 cm^3. The extra buoyant force equals the weight of the coin, so the mass of the coin is the mass of 387 cm^3 of water, which is 387 g. The cube's own floating does not enter the calculation because only the change in displacement is given.

Q7 (NCERT Exemplar, 1 mark)

An ideal fluid flows through a pipe of circular cross-section with diameters 2.5 cm and 3.75 cm in the two sections. What is the ratio of the velocities in the two sections?

Answer: By the equation of continuity, v is proportional to 1/A, and A is proportional to the square of the diameter. The velocity ratio is (3.75/2.5)^2 = (1.5)^2 = 2.25 = 9/4. The velocity is larger in the narrower section, so the ratio of the velocity in the narrow section to that in the wide section is 9:4.

Q8 (NCERT Exemplar, 1 mark)

The angles of contact of water-glass, ethyl alcohol-glass, mercury-glass and methyl iodide-glass are 0 degrees, 0 degrees, 140 degrees and 30 degrees. A capillary of glass is dipped into one of these liquids, and the meniscus is found to be convex. Which liquid is in the trough?

Answer: A convex meniscus means the liquid does not wet the glass, that is the contact angle is obtuse (greater than 90 degrees). Among the four, only mercury-glass has an obtuse contact angle of 140 degrees, so the liquid is mercury. Water, ethyl alcohol and methyl iodide all wet glass and give a concave meniscus.

Q9 (NCERT Exemplar, 3 marks)

The sap in trees rises in capillaries of radius 2.5 x 10^-5 m. The surface tension of sap is 7.28 x 10^-2 N m^-1 and the contact angle is zero. Can surface tension alone account for the supply of water to the top of a tall tree?

Answer: The capillary rise is h = 2T cos(theta)/(rho g r). Taking the density of sap as 1.0 x 10^3 kg m^-3, h = 2 x 7.28 x 10^-2/(1000 x 9.8 x 2.5 x 10^-5) = 0.594 m, which is about 60 cm. Real trees are several metres to tens of metres tall, so surface tension alone cannot raise water to the top. The main height in tall trees is supported by the transpiration pull, with the water column under tension, and capillarity only helps in the narrow xylem tubes.

Q10 (NCERT Exemplar, 2 marks)

An iceberg floats in sea water. The density of ice is 0.917 g cm^-3 and that of sea water is about 1.03 g cm^-3. What fraction of the iceberg's volume stays submerged?

Answer: For a floating body, the submerged fraction equals the ratio of the body density to the liquid density. The fraction is 0.917/1.03 = 0.89, so about 89 percent of the iceberg is under water and only about 11 percent shows above the surface. This small visible part is why an iceberg is a serious hazard for ships. The same rule explains why a loaded ship sinks deeper and why a hydrometer floats lower in a less dense liquid.