Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics) — page 6.
Q47 (2026)
The terminal velocity of a metallic ball of radius $6 \text{ mm}$ in a viscous fluid is $20 \text{ cm/s}$. The terminal velocity of another ball of same material and having radius $3 \text{ mm}$ in the same fluid will be ________ cm/s.
### Related Formula
$$v_T = \frac{2r^2 g}{9\eta} (\sigma - \rho)$$
### Core Logic
For balls of the same material falling through the same fluid, all terms in the terminal velocity equation except the radius $r$ are constant.
Therefore, $v_T \propto r^2$.
### Step 1: Setting up the Ratio
$$\frac{(v_T)_1}{(v_T)_2} = \left(\frac{r_1}{r_2}\right)^2$$
Substituting the given values:
$r_1 = 6 \text{ mm}$, $(v_T)_1 = 20 \text{ cm/s}$
$r_2 = 3 \text{ mm}$
### Step 2: Final Conclusion
$$\frac{20}{(v_T)_2} = \left(\frac{6}{3}\right)^2$$
$$\frac{20}{(v_T)_2} = 2^2 = 4$$
$$(v_T)_2 = \frac{20}{4} = 5 \text{ cm/s}$$
### Pattern Recognition
Terminal velocity is directly proportional to the square of the radius. Halving the radius ($6 \to 3$) will drop the terminal velocity by a factor of $1/4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q33 (2026)
Given below are two statements:
Statement I : Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.
Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.
In the light of the above statements, choose the correct answer from the options given below:
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is false but Statement II is true}$
### Related Formula
$$P = \frac{F}{A}, \quad \text{Surface Tension} \propto \Delta U$$
### Core Logic
According to Pascal's law, pressure exists at every point in a liquid at rest, not just at boundaries. Thus, Statement I is false.
For interior molecules, net cohesive forces are zero, whereas surface molecules possess excess potential energy leading to surface tension. Thus, Statement II is correct.
### Pattern Recognition
Sees: Fluid pressure definition + surface tension origin.
Shortcut: Recall Pascal's law (pressure everywhere) and surface energy excess.
Check: Matches option (4). ✓
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q34 (2026)
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises up to certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by ____%.
- $-1$
- $+3$
- $-3$
- $+1$
### Related Formula
$$h = \frac{2T \cos\theta}{\rho g r}$$
$$\frac{\Delta h}{h}\% = \frac{\Delta T}{T}\% - \frac{\Delta \rho}{\rho}\% - \frac{\Delta r}{r}\%$$
### Core Logic
Applying fractional error analysis to the capillary rise formula:
$$\frac{\Delta h}{h}\% = (-1\%) - (-1\%) - (-1\%)$$
$$\frac{\Delta h}{h}\% = -1 + 1 + 1 = +1\%$$
{{SOL_IMG1}}
### Step 1: Final Conclusion
The height of the liquid in the tube increases by $+1\%$.
### Pattern Recognition
Percentage error in capillary rise: $h \propto \frac{T}{\rho r}$.
Decreasing numerator $T$ by 1% reduces $h$ by 1% ($-1\%$).
Decreasing denominators $\rho$ and $r$ by 1% each increases $h$ by 1% twice ($+1\% + 1\%$).
Net change: $-1 + 1 + 1 = +1\%$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q45 (2026)
A small metallic sphere of diameter 2 mm and density $10.5 \, \mathrm{g/cm}^{3}$ is dropped in glycerine having viscosity 10 Poise and density $1.5 \, \mathrm{g/cm}^{3}$ respectively. The terminal velocity attained by the sphere is ____ cm/s.
($\pi = \frac{22}{7} \text{ and } g = 10 \, \mathrm{m/s^2}$)
- $2.0$
- $1.0$
- $3.0$
- $1.5$
### Related Formula
$$V_T = \frac{2r^2 g}{9\eta} (\rho_b - \rho_{\ell})$$
### Core Logic
Given data:
Radius $r = 1 \, \mathrm{mm} = 0.1 \, \mathrm{cm}$.
Density of sphere $\rho_b = 10.5 \, \mathrm{g/cm}^3$.
Density of liquid $\rho_{\ell} = 1.5 \, \mathrm{g/cm}^3$.
Viscosity $\eta = 10 \text{ Poise} = 10 \, \mathrm{g/cm\cdot s}$ (since 1 Poise = 1 CGS unit of viscosity).
Gravity $g = 10 \, \mathrm{m/s}^2 = 1000 \, \mathrm{cm/s}^2$ (Wait, standard gravity for CGS is $980$, but the problem specifies $g = 10 \, \mathrm{m/s}^2 = 1000 \, \mathrm{cm/s}^2$).
### Step 1: Calculation in CGS Units
$$V_T = \frac{2}{9} \cdot \frac{(0.1)^2 \times 1000}{(10)} \cdot (10.5 - 1.5)$$
$$V_T = \frac{2}{9} \cdot \frac{0.01 \times 1000}{10} \cdot (9)$$
$$V_T = 2 \cdot \frac{10}{10} \cdot 1$$
$$V_T = 2 \, \mathrm{cm/s}$$
### Pattern Recognition
Watch the units closely. Poise is the CGS unit for viscosity, so converting all values to CGS (cm, g, s) makes the calculation direct and prevents factor-of-10 errors common when mixing SI and CGS.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q50 (2026)
Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ____ cm/s.
### Related Formula
$$V_T = \frac{2r^2 g}{9\eta} (\rho - \sigma)$$
$V_T \propto r^2$
### Core Logic
{{SOL_IMG1}}
When 64 small drops (radius $R_1$) coalesce to form a bigger drop (radius $R_2$), mass and volume are conserved.
$$64 \left(\frac{4}{3} \pi R_1^3\right) = \frac{4}{3} \pi R_2^3$$
$$R_2^3 = 64 R_1^3 \implies R_2 = 4 R_1$$
{{SOL_IMG2}}
### Step 1: Terminal Velocity Ratio
Since terminal velocity $V_T \propto R^2$:
$$\frac{(V_T)_1}{(V_T)_2} = \left( \frac{R_1}{R_2} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$$
Given $(V_T)_1 = 10 \text{ cm/s}$:
$$\frac{10}{(V_T)_2} = \frac{1}{16}$$
$$(V_T)_2 = 160 \text{ cm/s}$$
### Pattern Recognition
When $N$ identical drops coalesce, the new radius is $N^{1/3}$ times the old radius. Consequently, the new terminal velocity scales precisely as $N^{2/3}$. For $N=64$, $64^{2/3} = 16$. Just multiply initial $v$ by 16.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
