PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 9 · JEE Main

Mechanical Properties of Fluids

Chapter-wise practice and revision for JEE Main: concise notes on pressure, buoyancy, Bernoulli's principle, viscosity and surface tension, concept figures, and 45 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

45 selected questions5 topics5 concept figuresPreview · first 3 of 45 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 134 previous-year JEE questions in the PiTheory bank

Surface tension48
Viscosity30
Bernoulli26
Pressure19
Buoyancy11

Contents

  1. 1Pressure, Pascal's law and hydraulic machinesQ1–Q6
  2. 2Archimedes' principle and floating bodiesQ7–Q11
  3. 3Equation of continuity and Bernoulli's principleQ12–Q20
  4. 4Viscosity, Stokes' law and terminal velocityQ21–Q30
  5. 5Surface tension, capillarity, drops and bubblesQ31–Q45
  6. ✓Last-day checklist
01

Topic 1 · Q1–Q6

Pressure, Pascal's law and hydraulic machines

19 of the 134 JEE PYQs in the bank test this as their main idea

Look first

F₁loadA₁A₂same pressure: F₁/A₁ = F₂/A₂
Pascal's law: pressure is passed on undiminished.

Quick notes

$$P=P_0+\rho gh,\qquad \frac{F_1}{A_1}=\frac{F_2}{A_2}\ \text{(Pascal)}$$

Pressure at a depth depends only on the depth, not on the shape of the vessel. Points at the same level in the same connected liquid at rest have the same pressure. Gauge pressure $=P-P_0$.

SituationResult
Hydraulic liftforce $\times\dfrac{A_2}{A_1}=\left(\dfrac{D}{d}\right)^2$; work in $=$ work out (small piston moves farther)
Force on a vertical wall, width $b$, depth $H$average pressure $\times$ area $=\tfrac12\rho gH\cdot bH$
Liquid in an accelerating containersurface tilts: $\tan\theta=\dfrac ag$
Liquid rotating at $\omega$paraboloid surface; rise at radius $R$: $\dfrac{\omega^2R^2}{2g}$
Two vessels (area $S$) joined at the bottomlevel $\dfrac{x_1+x_2}{2}$; energy lost $\tfrac14\rho gS(x_1-x_2)^2$
U-tube with two liquidsequate pressures at the lower interface level: $\rho_1h_1=\rho_2h_2$
Traps. Add $P_0$ only when the question asks for absolute pressure or the force on a surface whose other side is not open to the atmosphere (if air presses on both sides, $P_0$ cancels). Pressure is a scalar: it acts equally in all directions.

Practice: first 3 of 6 questions

Q1JEE Main 2019 (10 Apr, Shift 2)Easy

A submarine experiences a pressure of $5.05\times10^6\ \text{Pa}$ at a depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08\times10^6\ \text{Pa}$. Then $d_2 - d_1$ is approximately (density of water $= 10^3\ \text{kg/m}^3$ and acceleration due to gravity $= 10\ \text{m s}^{-2}$):

  1. A$600\ \text{m}$
  2. B$500\ \text{m}$
  3. C$300\ \text{m}$
  4. D$400\ \text{m}$
Answer and solution

Answer: Option C: $300\ \text{m}$

Concept. In a liquid at rest, pressure grows linearly with depth: $P=P_0+\rho gd$. The difference between two depths depends only on the depth difference.

Step 1. Pressure difference. $\Delta P=8.08\times10^6-5.05\times10^6=3.03\times10^6$ Pa.

Step 2. Depth difference. $\Delta P=\rho g\,\Delta d\Rightarrow\Delta d=\dfrac{3.03\times10^6}{10^3\times10}=303$ m.

Answer. About $300$ m (Option C).

Point to note. $\Delta P=\rho g\,\Delta d$.

Q2JEE Main 2021 (24 Feb, Shift 1)Medium

A hydraulic press can lift 100 kg when a mass $m$ is placed on the smaller piston. It can lift ______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass $m$ on the smaller piston.

Numerical answer type

Answer and solution

Answer: 25600

Concept. Pascal's law: the pressure added at the small piston reaches the large piston unchanged. So $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$, i.e. the load lifted is $F_2=F_1\dfrac{A_2}{A_1}=F_1\left(\dfrac{D}{d}\right)^2$.

Step 1. The new area ratio. Large diameter $\times4$ (area $\times16$); small diameter $\div4$ (area $\div16$). The ratio $\dfrac{A_2}{A_1}$ grows by $16\times16=256$.

Step 2. New load (same $m$ on the small piston): $100\times256=25\,600$ kg.

Answer. $25\,600$.

Point to note. Load $\propto(D/d)^2$; both changes multiply.

Q3JEE Main 2025 (29 Jan, Shift 1)Easy

In a hydraulic lift, the surface area of the input piston is $6\ \text{cm}^2$ and that of the output piston is $1500\ \text{cm}^2$. If a $100\ \text{N}$ force is applied to the input piston to raise the output piston by $20\ \text{cm}$, then the work done is ______ kJ.

Numerical answer type

Answer and solution

Answer: 5

Concept. A hydraulic lift multiplies force but not work. The output force is $F_2=F_1\dfrac{A_2}{A_1}$, and the work done is $F_2\times$ the rise of the output piston (equal to $F_1\times$ the input piston's longer stroke).

Step 1. Output force. $F_2=100\times\dfrac{1500}{6}=25\,000$ N.

Step 2. Work. $W=25\,000\times0.20=5000$ J $=5$ kJ.

Answer. $5$.

Point to note. Output force $\times$ distance; work is not multiplied.

42 more questions and 4 more topics in the full chapter

  1. 2Archimedes' principle and floating bodies5 questions with solutions
  2. 3Equation of continuity and Bernoulli's principle9 questions with solutions
  3. 4Viscosity, Stokes' law and terminal velocity10 questions with solutions
  4. 5Surface tension, capillarity, drops and bubbles15 questions with solutions
  5. ✓Last-day checklist

The complete Mechanical Properties of Fluids chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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