Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics).
Q4 (2025)
Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension, the surface energy released in this process is:
- $4\pi r^2 T \left[2 - 2^{\frac{2}{3}}\right]$
- $4\pi r^2 T \left[2 - 2^{\frac{1}{3}}\right]$
- $4\pi r^2 T \left[1 + \sqrt{2}\right]$
- $4\pi r^2 T \left[\sqrt{2} - 1\right]$
### Related Formula
1. Surface Energy:
$$U = T \cdot A = T \cdot (4\pi R^2)$$
2. Conservation of Volume during coalescence of drops:
$$2 \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$$
### Core Logic
When two drops of radius $r$ coalesce into a single larger drop of radius $R$, volume is conserved:
$$R^3 = 2r^3 \implies R = 2^{1/3} r$$
- Initial surface area of the two separate drops:
$$A_i = 2 \times 4\pi r^2 = 8\pi r^2$$
- Final surface area of the combined single drop:
$$A_f = 4\pi R^2 = 4\pi (2^{1/3} r)^2 = 4\pi r^2 2^{2/3}$$
- Surface energy released:
$$\Delta E = U_i - U_f = T(A_i - A_f)$$
$$\Delta E = T \left( 8\pi r^2 - 4\pi r^2 2^{2/3} \right) = 4\pi r^2 T \left[ 2 - 2^{2/3} \right]$$
### Pattern Recognition
Sees: Coalescence of $N$ identical drops.
Trap: Forgetting to conserve volume first, or confusing initial and final surface areas.
Shortcut: Energy released when $N$ drops coalesce into one big drop is:
$$\Delta E = 4\pi r^2 T \left[ N - N^{2/3} \right]$$
Here, substituting $N = 2$ directly gives $4\pi r^2 T \left[ 2 - 2^{2/3} \right]$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q21 (2025)
A vessel with square cross-section and height of $6\mathrm{~m}$ is vertically partitioned. A small window of $100\mathrm{~cm}^2$ with hinged door is fitted at a depth of $3\mathrm{~m}$ in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density $1.5\times 10^{3}\mathrm{~kg/m}^{3}$. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity $= 10\mathrm{~m/s}^2$)
### Related Formula
$$P = P_0 + \rho g h$$
$$F = \Delta P \cdot A$$
### Core Logic
Let's analyze the pressure acting on the window at depth $h = 3\mathrm{~m}$ from both sides:
- Water side (density $\rho_w = 1.0 \times 10^3\mathrm{~kg/m}^3$):
$$P_w = P_0 + \rho_w g h$$
- Denser liquid side (density $\rho_l = 1.5 \times 10^3\mathrm{~kg/m}^3$):
$$P_l = P_0 + \rho_l g h$$
The net pressure difference pushing on the window is:
$$\Delta P = P_l - P_w = (\rho_l - \rho_w) g h$$
$$\Delta P = (1.5 \times 10^3 - 1.0 \times 10^3) \times 10 \times 3 = 500 \times 30 = 15000\mathrm{~N/m}^2$$
The window has area $A = 100\mathrm{~cm}^2 = 100 \times 10^{-4}\mathrm{~m}^2 = 10^{-2}\mathrm{~m}^2$.
To keep the door from opening, we must apply a force balancing the pressure difference:
$$F = \Delta P \cdot A = 15000 \times 10^{-2} = 150\mathrm{~N}$$$
### Step 1: Final Conclusion
The required force is $150\mathrm{~N}$.
### Pattern Recognition
For a vertical interface between two fluids, the net pressure difference is simply given by $\Delta P = \Delta \rho \cdot g h$. The ambient atmospheric pressure $P_0$ cancels out since it acts on both sides of the window.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q153 (2025)
Two cylindrical vessels of equal cross sectional area of $2\mathrm{~m}^{2}$ contain water upto height $10\mathrm{~m}$ and $6\mathrm{~m}$, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is :
(Density of water is $10^{3}\mathrm{~kg/m}^{3}$ and $g=10\mathrm{~m/s}^{2}$)
- $1 \times 10^{5}\mathrm{~J}$
- $4 \times 10^{4}\mathrm{~J}$
- $6 \times 10^{4}\mathrm{~J}$
- $8 \times 10^{4}\mathrm{~J}$
### Related Formula
The gravitational potential energy $U$ of a liquid column of mass $m$ and height $h$ is evaluated relative to its bottom by placing its total mass at its center of mass ($h/2$):
$$U = m g \left(\frac{h}{2}\right) = (\rho A h) g \left(\frac{h}{2}\right) = \frac{1}{2} \rho A g h^2$$
Work done by the force of gravity ($W$) equals the negative change in potential energy:
$$W = -\Delta U = U_i - U_f$$
### Core Logic
Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at:
$$h_f = \frac{10 + 6}{2} = 8\mathrm{~m}$$
### Step 1: Calculate Initial Potential Energy ($U_i$)
Let the reference level $U=0$ be at the bottom:
$$U_i = U_1 + U_2 = \frac{1}{2} \rho A g h_1^2 + \frac{1}{2} \rho A g h_2^2$$
$$U_i = \frac{1}{2} \rho A g \left(10^2 + 6^2\right) = \frac{1}{2} \rho A g (100 + 36) = 68 \rho A g$$
{{SOL_IMG1}}
### Step 2: Calculate Final Potential Energy ($U_f$)
Both vessels equalize to $h_f = 8\mathrm{~m}$:
$$U_f = 2 \times \left[ \frac{1}{2} \rho A g h_f^2 \right] = \rho A g (8^2) = 64 \rho A g$$
### Step 3: Work Done by Gravity ($W$)
$$W = U_i - U_f = 68 \rho A g - 64 \rho A g = 4 \rho A g$$
Substitute the given values ($
ho = 10^3\mathrm{~kg/m}^3$, $A = 2\mathrm{~m}^2$, $g = 10\mathrm{~m/s}^2$):
$$W = 4 \times 10^3 \times 2 \times 10 = 8 \times 10^4\mathrm{~J}$$$
### Pattern Recognition
For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2$. Applying this directly:
$$\Delta U = \frac{1}{4} \times 10^3 \times 2 \times 10 \times (10 - 6)^2 = 5000 \times 16 = 8 \times 10^4\mathrm{~J}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q12 (2025)
A solid steel ball of diameter $3.6\mathrm{~mm}$ acquired terminal velocity $2.45\times10^{-2}\mathrm{~m/s}$ while falling under gravity through an oil of density $925\mathrm{~kg~m}^{-3}$. Take density of steel as $7825\mathrm{~kg~m}^{-3}$ and g as $9.8\mathrm{~m/s}^{2}$. The viscosity of the oil in SI unit is :
- 2.18
- 2.38
- 1.68
- 1.99
### Related Formula
Terminal velocity $v_t$ of a spherical body falling through a viscous fluid is given by Stokes' Law:
$$v_t = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{\eta}$$
Hence, the viscosity coefficient $\eta$ is:
$$\eta = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{v_t}$$
### Core Logic
Given parameters:
- Diameter $d = 3.6\mathrm{~mm} \Rightarrow$ radius $r = 1.8\mathrm{~mm} = 1.8 \times 10^{-3}\mathrm{~m}$
- Terminal velocity $v_t = 2.45 \times 10^{-2}\mathrm{~m/s}$
- Liquid density $\rho_{\text{fluid}} = 925\mathrm{~kg/m}^3$
- Steel density $\rho_{\text{body}} = 7825\mathrm{~kg/m}^3$
- Acceleration due to gravity $g = 9.8\mathrm{~m/s}^2$
### Step 1: Substitute parameters into formula
$$\eta = \frac{2}{9} \times \frac{(1.8 \times 10^{-3})^2 \times 9.8 \times (7825 - 925)}{2.45 \times 10^{-2}}$$
$$\eta = \frac{2}{9} \times \frac{3.24 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$$
### Step 2: Solve the numeric calculation
$$\eta = \frac{2 \times 0.36 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$$
$$\eta = \frac{0.72 \times 9.8 \times 6900 \times 10^{-6}}{2.45 \times 10^{-2}}$$
$$\eta = \frac{48700.8 \times 10^{-6}}{2.45 \times 10^{-2}} = \frac{0.0487008}{0.0245} \approx 1.99\mathrm{~Pa\cdot s}$$$
### Pattern Recognition
Be careful when converting diameter to radius ($r = d/2$). Always ensure all numerical parameters are converted cleanly to SI base units (meters, kilograms, seconds) before applying Stokes' formula.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q22 (2025)
The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is $n$ times the volume of the bubble B, then the value of $n$ is ________.
### Related Formula
Excess pressure inside a soap bubble in air (which has two liquid-gas interfaces) is given by:
$$\Delta P = \frac{4T}{R} \Rightarrow R \propto \frac{1}{\Delta P}$$
The volume $V$ of a spherical bubble of radius $R$ is:
$$V = \frac{4}{3}\pi R^3 \Rightarrow V \propto R^3 \propto \left(\frac{1}{\Delta P}\right)^3$$
### Core Logic
Given state:
$$\Delta P_A = \frac{1}{2} \Delta P_B \Rightarrow \frac{\Delta P_B}{\Delta P_A} = 2$$
### Step 1: Calculate the ratio of radii
$$\frac{R_A}{R_B} = \frac{\Delta P_B}{\Delta P_A} = 2$$
### Step 2: Calculate volume scaling parameter ($n$)
$$V_A = n V_B \Rightarrow n = \frac{V_A}{V_B} = \left(\frac{R_A}{R_B}\right)^3 = (2)^3 = 8$$
### Pattern Recognition
Volume scale factors depend on the cube of linear scale factors ($V \propto R^3$). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by $2^3 = 8$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
