Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics) — page 5.
Q52 (2024)
A big drop is formed by coalescing $1000$ small identical drops of water. If $\mathrm{E}_1$ be the total surface energy of $1000$ small drops of water and $\mathrm{E}_2$ be the surface energy of single big drop of water, the $\mathrm{E}_1:\mathrm{E}_2$ is $x:1$ where $x =$
### Related Formula
$$V_{\text{initial}} = V_{\text{final}}$$
$E = S \times A$
### Core Logic
When small drops coalesce to form a large drop, the total volume is conserved.
$$1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$
$$R^3 = 1000 r^3 \implies R = 10r$$
### Step 1: Calculate Surface Energies
The total surface energy of $1000$ small drops ($\mathrm{E}_1$) is:
$$\mathrm{E}_1 = 1000 \times 4\pi r^2 \times S$$
The surface energy of the single big drop ($\mathrm{E}_2$) is:
$$\mathrm{E}_2 = 4\pi R^2 \times S = 4\pi (10r)^2 \times S = 100 \times 4\pi r^2 \times S$$
### Step 2: Find the Ratio
$$\frac{\mathrm{E}_1}{\mathrm{E}_2} = \frac{1000}{100} = \frac{10}{1}$$
Thus, the ratio is $10:1$, which means $x = 10$.
### Pattern Recognition
When $N$ droplets merge to form one big drop, the radius scales as $R = N^{1/3} r$. The ratio of total initial surface energy to final surface energy is $N^{1/3}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q47 (2024)
A small spherical ball of radius $r$, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius $2r$, falling through the same viscous medium will have terminal velocity:
- $\frac{v}{2}$
- $\frac{v}{4}$
- $4v$
- $2v$
### Related Formula
At terminal velocity, downward force equals upward drag (assuming negligible buoyancy):
$$Mg = 6\pi \eta r v$$
$$v = \frac{Mg}{6\pi \eta r}$$
### Core Logic
Since the density of the medium is negligible, we ignore buoyant forces.
The mass $M$ of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower).
### Step 1: Setup Proportionality
Since $M$, $g$, and $\eta$ are all constants:
$$v \propto \frac{1}{r}$$
### Step 2: Evaluating the Ratio
For the second ball, $r' = 2r$.
Therefore, the new terminal velocity $v'$ is:
$$v' = v \times \left(\frac{r}{r'}\right) = v \times \left(\frac{r}{2r}\right) = \frac{v}{2}$$
### Pattern Recognition
Read the constraints carefully. Usually, questions keep material density uniform ($v \propto r^2$). However, this specifically says "same mass". This shifts the formula dependency from $v \propto r^2$ entirely to $v \propto 1/r$ because $M$ acts as a constant numerator.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q1996 (2024)
A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?
### Related Formula
$$mg - F_B - F_v = ma$$
$$F_v = 6\pi\eta r v$$
### Core Logic
{{SOL_IMG1}}
When dropped, three forces act on the ball: Gravity downwards, Buoyant force upwards, and Viscous drag upwards.
$$mg - F_B - F_v = m \frac{dv}{dt}$$
$$\left(\rho \frac{4}{3}\pi r^3\right)g - \left(\rho_L \frac{4}{3}\pi r^3\right)g - 6\pi\eta rv = m \frac{dv}{dt}$$
Let $\frac{4\pi r^3 g(\rho - \rho_L)}{3m} = K_1$ and $\frac{6\pi\eta r}{m} = K_2$.
$$\frac{dv}{dt} = K_1 - K_2 v$$
Integrating from $t=0, v=0$:
$$\int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt$$
$$-\frac{1}{K_2} \ln \left(\frac{K_1 - K_2 v}{K_1}\right) = t$$
$$v = \frac{K_1}{K_2} \left( 1 - e^{-K_2 t} \right)$$
This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity $V_T = K_1 / K_2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties Of Fluids
Q2074 (2026)
Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are $6\text{ cm}^{2}$ and $3\text{ cm}^{2}$ respectively. The rate of flow will be ____ $\text{cm}^{3}$/s. (take g = 10 $\text{m/s}^{2}$)
{{IMG1}}
- $\frac{200}{\sqrt{3}}$
- $200\sqrt{6}$
- $200\sqrt{3}$
- $100\sqrt{3}$
### Related Formula
$$A_1 V_1 = A_2 V_2$$
$$P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2$$
$$P_1 - P_2 = \rho g h$$
### Core Logic
From the continuity equation between A and B:
$$A_{A}V_{A} = A_{B}V_{B} \implies 6V_{A} = 3V_{B} \implies V_{B} = 2V_{A}$$
Applying Bernoulli's equation between A and B for a horizontal pipe:
$$P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2}$$
$$P_{A} - P_{B} = \frac{1}{2}\rho(V_{B}^{2} - V_{A}^{2})$$
Since the height difference is $h = 5\text{ cm} = 0.05\text{ m}$, the pressure difference is $\rho gh$.
$$\rho g \times 0.05 = \frac{1}{2}\rho( (2V_A)^2 - V_A^2 )$$
$$g \times 0.05 = \frac{1}{2}(3V_{A}^{2})$$
### Step 1: Calculate Velocity and Volume Flow Rate
Solving for $V_A$:
$$V_{A} = \sqrt{\frac{2 \times 10 \times 0.05}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\text{ m/s}$$
$$V_A = \frac{100}{\sqrt{3}}\text{ cm/s}$$
Volume flow rate $Q = A_{A} V_{A}$:
$$Q = 6\text{ cm}^2 \times \frac{100}{\sqrt{3}}\text{ cm/s} = \frac{600}{\sqrt{3}} = 200\sqrt{3}\text{ cm}^{3}\text{/s}$$
### Pattern Recognition
In a horizontal Venturi meter, substituting $V_2 = V_1 (A_1/A_2)$ directly into $\rho g h = \frac{1}{2}\rho (V_2^2 - V_1^2)$ is the standard path. Working in CGS vs MKS units requires careful tracking; convert $V_A$ to cm/s before multiplying by $A_A$ in cm².
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q38 (2026)
Surface tension of two liquids (having same densities), $T_{1}$ and $T_{2}$, are measured using capillary rise method utilizing two tubes with inner radii of $r_{1}$ and $r_{2}$ where $r_{1} > r_{2}$. The measured liquid heights in these tubes are $h_{1}$ and $h_{2}$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights $h_{1}$ and $h_{2}$ and surface tensions $T_{1}$ and $T_{2}$ satisfy the relation :
- $h_1 < h_2 \text{ and } T_1 = T_2$
- $h_1 = h_2 \text{ and } T_1 = T_2$
- $h_1 > h_2 \text{ and } T_1 = T_2$
- $h_1 > h_2 \text{ and } T_1 < T_2$
### Related Formula
$$h = \frac{2T\cos\theta}{\rho g r}$$
### Core Logic
Since we are assessing surface tension $T$ as a property of the liquids, the question intends to compare liquids that are identical (implying $T_1 = T_2$ and same density).
{{SOL_IMG1}}
By Jurin's Law, for identical liquids, the height of capillary rise is inversely proportional to the radius of the tube:
$$h \propto \frac{1}{r}$$
### Step 1: Final Conclusion
Given $r_1 > r_2$, the inverse proportionality directly implies that the liquid will rise less in the wider tube.
Therefore, $h_1 < h_2$ and naturally $T_1 = T_2$ for the intrinsic property.
### Pattern Recognition
Wider tubes ($r$ large) cause lower capillary rises ($h$ small). $h \propto 1/r$ is a foundational inverse relationship in capillary action.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
