Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics) — page 3.
Q8 (2025)
A capillary tube of radius 0.1 mm is partly dipped in water (surface tension $70~dyn/cm$ and glass water contact angle $simeq0^{circ}$) with $30^{circ}$ inclined with vertical.
The length of water risen in the capillary is cm.
(Take $g=9.8~m/s^{2}$) [cite: 83, 84]
- $\frac{82}{5}$ [cite: 85]
- $\frac{57}{2}$ [cite: 88]
- $\frac{71}{5}$ [cite: 86]
- $\frac{68}{5}$ [cite: 88]
### Related Formula
$$h = \frac{2T\cos\theta}{\rho g r}$$ [cite: 694]
### Core Logic
First, calculate the vertical height $h$ using CGS units: [cite: 694]
* $T = 70\ \text{dyn/cm}$ [cite: 83]
* $\cos\theta = \cos 0^{\circ} = 1$ [cite: 83]
* $\rho = 1\ \text{g/cm}^3$ [cite: 694]
* $g = 980\ \text{cm/s}^2$ [cite: 694]
* $r = 0.1\ \text{mm} = 10^{-2}\ \text{cm}$ [cite: 83, 694]
$$h = \frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} = \frac{140}{9.8} = \frac{100}{7}\ \text{cm}$$ [cite: 694]
Since the tube is inclined at $30^{\circ}$ to the vertical, the angle it makes with the horizontal is $60^{\circ}$[cite: 83, 694]. The relationship between vertical height $h$ and slant length $l$ is: [cite: 694]
$$\sin 60^{\circ} = \frac{h}{l} \implies l = \frac{h}{\sin 60^{\circ}} = \frac{2h}{\sqrt{3}}$$ [cite: 694]
$$l = \frac{100}{7} \times \frac{2}{\sqrt{3}} = \frac{200}{7\sqrt{3}} \approx 16.49\ \text{cm}$$ [cite: 696, 697, 698]
Checking option values, $\frac{82}{5} = 16.4\ \text{cm}$, which is closest[cite: 85, 698].
### Pattern Recognition
When a capillary tube is tilted, the vertical height of the fluid column remains constant to maintain hydrostatic pressure balance[cite: 694]. Thus, the length of the liquid along the slant always scales as $l = \frac{h}{\cos\alpha}$ where $\alpha$ is the tilt angle relative to the vertical line[cite: 83, 694].
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q1132 (2025)
The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-
- 15 J
- 10 J
- 20 J
- 5 J
### Related Formula
The work done $W$ in expanding surface area with surface tension $S$ is given by:
$$W = S \cdot \Delta A$$
### Core Logic
By volume conservation during splitting :
$$\frac{4}{3}\pi R^{3} = n \left(\frac{4}{3}\pi r^{3}\right) \implies r = \frac{R}{n^{1/3}}$$
Total change in area gives work expression:
$$W = S \left(n \cdot 4\pi r^{2} - 4\pi R^{2}\right) = 4\pi R^{2}S \left(n^{1/3} - 1\right)$$
### Step 1: Set up Proportions
For $n = 27$ drops:
$$W = 4\pi R^{2}S (27^{1/3} - 1) = 4\pi R^{2}S (3 - 1) = 2(4\pi R^{2}S) = 10\text{ J}$$
$$4\pi R^{2}S = 5\text{ J}$$
For $n = 64$ drops:
$$W' = 4\pi R^{2}S (64^{1/3} - 1) = 4\pi R^{2}S (4 - 1) = 3(4\pi R^{2}S)$$
$$W' = 3 \times 5 = 15\text{ J}$$
### Pattern Recognition
Work scales scaling-wise linearly with the key multiplier index $(n^{1/3} - 1)$. Taking ratios between targets directly avoids calculations of constants.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q3 (2025)
An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density $1000\text{ kg/m}^{3}$ If the pressure inside the bubble is 2100 N/m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $g=10\text{ m/s}^{2})$
- 0.02
- 0.1
- 0.25
- 0.04
### Related Formula
The absolute pressure inside an air bubble submerged in a liquid is given by :
$$P_{\text{in}} = P_{0} +
ho gh + \frac{2T}{R}$$
where $P_0$ is atmospheric pressure, $
ho$ is the liquid density, $g$ is gravity, $h$ is depth, $T$ is surface tension, and $R$ is the radius.
### Core Logic
We are given that the difference between the inside pressure and atmospheric pressure is $2100\text{ N/m}^2$:
$$P_{\text{in}} - P_{0} =
ho gh + \frac{2T}{R} = 2100$$
### Step 1: Numerical Evaluation
Substitute the given values into the relation :
$$
ho gh = 1000 \times 10 \times 0.20 = 2000\text{ N/m}^2$$
Now find the excess pressure from surface tension:
$$\frac{2T}{R} = 2100 - 2000 = 100\text{ N/m}^2$$
$$T = \frac{100 \times R}{2} = 50 \times (0.1 \times 10^{-2}) = 0.05\text{ N/m}$$
### Pattern Recognition
Total inside pressure accounts for both the hydrostatic pressure of the fluid column ($
ho gh$) and the spherical geometry boundary constraint pressure ($\frac{2T}{R}$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q18 (2025)
A $400\mathrm{g}$ solid cube having an edge of length $10\mathrm{cm}$ floats in water. How much volume of the cube is outside the water?
(Given: density of water $= 1000\mathrm{kg}\mathrm{m}^{-3}$ )
- $1400\mathrm{cm}^3$
- $4000\mathrm{cm}^3$
- $400\mathrm{cm}^3$
- $600~\mathrm{cm}^3$
### Related Formula
By the law of flotation, the weight of a floating body must exactly balance the buoyant force exerted by the displaced fluid volume:
$$M \cdot g = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g$$
### Core Logic
Given parameters:
* Mass of the cube, $M = 400 \text{ g} = 0.4 \text{ kg}$
* Total volume of the cube, $V_{\text{total}} = (10 \text{ cm})^3 = 1000 \text{ cm}^3 = 10^{-3} \text{ m}^3$
* Density of water, $\rho_{\text{water}} = 1000 \text{ kg/m}^3$
Equating weight to buoyant force to find the submerged volume $V_d$ :
$$0.4 = 1000 \times V_{\text{submerged}}$$
$$V_{\text{submerged}} = \frac{0.4}{1000} = 4 \times 10^{-4} \text{ m}^3 = 400 \text{ cm}^3 \quad \text{}$$
Calculate the volume remaining outside the water surface :
$$V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}}$$
$$V_{\text{outside}} = 1000 \text{ cm}^3 - 400 \text{ cm}^3 = 600 \text{ cm}^3 \quad \text{}$$
### Pattern Recognition
The fraction of a floating body's volume that is submerged equals the ratio of the body's density to the fluid's density: $\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$. Here, the cube's effective density is $0.4 \text{ g/cm}^3$, meaning $40\%$ is submerged and $60\%$ stays outside.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q1299 (2025)
In a hydraulic lift, the surface area of the input piston is $6\mathrm{cm}^2$ and that of the output piston is $1500\mathrm{cm}^2$ . If $100\mathrm{N}$ force is applied to the input piston to raise the output piston by $20\mathrm{cm}$ , then the work done is ________ kJ.
### Related Formula
$$W = F_1 \cdot s_1 = F_2 \cdot s_2$$
### Core Logic
By conservation of liquid volume displacement during output piston elevation :
$$A_1 \cdot s_1 = A_2 \cdot s_2 \implies 6 \cdot s_1 = 1500 \cdot 20 \implies s_1 = 5000\text{ cm} = 50\text{ m}$$
Work performed on input boundary matches :
$$W = F_1 \cdot s_1 = 100\text{ N} \cdot 50\text{ m} = 5000\text{ J} = 5\text{ kJ}$$
Alternatively via output force profile calculation: {{SOL_IMG1}}
$$F_2 = F_1 \frac{A_2}{A_1} = 100 \frac{1500}{6} = 25000\text{ N}$$
$$W = F_2 \cdot s_2 = 25000 \cdot 0.2 = 5000\text{ J} = 5\text{ kJ}$$
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
