Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics) — page 2.
Q22 (2025)
A sample of a liquid is kept at $1\mathrm{~atm}$. It is compressed to $5\mathrm{~atm}$ which leads to change of volume of $0.8\mathrm{~cm}^{3}$. If the bulk modulus of the liquid is $2\mathrm{~GPa}$, the initial volume of the liquid was ________ litre. (Take $1\mathrm{~atm} = 10^{5}\mathrm{~Pa}$)
### Related Formula
$$B = -\frac{\Delta P}{\Delta V / V} \implies V = B \frac{-\Delta V}{\Delta P}$$
where,
$B$ = Bulk modulus of liquid
$\Delta P$ = change in pressure
$\Delta V$ = change in volume
$V$ = initial volume
### Core Logic
Given parameters:
- Initial pressure, $P_i = 1\mathrm{~atm}$
- Final pressure, $P_f = 5\mathrm{~atm}$
- Change in pressure, $\Delta P = P_f - P_i = 4\mathrm{~atm} = 4 \times 10^5\mathrm{~Pa}$
- Change in volume, $\Delta V = -0.8\mathrm{~cm}^{3} = -0.8 \times 10^{-6}\mathrm{~m}^{3}$
- Bulk modulus, $B = 2\mathrm{~GPa} = 2 \times 10^9\mathrm{~Pa}$
### Step 1: Compute Initial Volume
Substitute the parameters into the formula:
$$V = 2 \times 10^9 \times \frac{0.8 \times 10^{-6}}{4 \times 10^5}$$
$$V = \frac{1.6 \times 10^3}{4 \times 10^5} = 0.4 \times 10^{-2} = 4 \times 10^{-3}\mathrm{~m}^{3}$$
Convert cubic meters to litres:
$$1\mathrm{~m}^3 = 1000\mathrm{~litres}$$
$$V = 4 \times 10^{-3} \times 1000 = 4\mathrm{~litres}$$
Thus, the initial volume was $4\mathrm{~litres}$.
### Pattern Recognition
Sees: Bulk modulus definition calculation.
Trap: Watch out for unit conversions: $1\mathrm{~GPa} = 10^9\mathrm{~Pa}$, and $1\mathrm{~cm}^3 = 10^{-6}\mathrm{~m}^3$. Finally, express the answer in Litres, where $1\mathrm{~L} = 10^{-3}\mathrm{~m}^3$. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q2 (2025)
In the experiment for measurement of viscosity $\eta$ of given liquid with a ball having radius $R$ , consider following statements.
A. Graph between terminal velocity V and R will be a parabola
B. The terminal velocities of different diameter balls are constant for a given liquid.
C. Measurement of terminal velocity is dependent on the temperature.
D. This experiment can be utilized to assess the density of a given liquid.
E. If balls are dropped with some initial speed, the value of $\eta$ will change.
Choose the correct answer from the options given below:
- $\text{B, D and E only}$
- $\text{A, C and D only}$
- $\text{C, D and E only}$
- $\text{A, B and E only}$
### Related Formula
$$\mathrm{V}_{\mathrm{T}} = \frac{2}{9} \mathrm{R}^{2} \frac{\mathrm{g}}{\eta} (\mathrm{d} - \rho)$$
### Core Logic
Let us check each statement sequentially:
Statement A: Since $\mathrm{V}_{\mathrm{T}} \propto \mathrm{R}^2$, the graph between terminal velocity $\mathrm{V}$ and radius $\mathrm{R}$ is a quadratic curve, i.e., a parabola. (True)
Statement B: Different diameters imply different radii, hence terminal velocity will change. It is not constant. (False)
Statement C: Viscosity of a liquid is heavily temperature-dependent, hence $\mathrm{V}_{\mathrm{T}}$ changes with temperature. (True)
Statement D: Knowing the parameters allows calculation of liquid density $\rho$. (True)
Statement E: $\eta$ is a material property of the fluid and does not change based on the initial dropping speed of the ball. (False)
### Step 1: Final Conclusion
Statements A, C, and D are correct, which matches option (2).
### Pattern Recognition
Terminal velocity problems rely entirely on parsing the proportional relations hidden in Stokes' Law formulation. Note that viscosity coefficient is an intrinsic characteristic parameter independent of kinematics.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q3 (2025)
Consider following statements:
A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid.
B. As the temperature of liquid rises, the coefficient of viscosity increases.
C. As the temperature of gas increases, the coefficient of viscosity increases.
D. The onset of turbulence is determined by Reynold's number.
E. In a steady flow two stream lines never intersect.
Choose the correct answer from the options given below:
- $\text{A, D, E only}$
- $\text{C, D, E only}$
- $\text{B, C, D only}$
- $\text{A, B, C only}$
### Core Logic
Let's audit each fluid mechanics assertion:
Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward.
Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease.
Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange, increasing viscosity.
Statement D: Correct. Critical velocity and turbulence parameters are completely defined by the dimensionless Reynolds number.
Statement E: Correct. If streamlines intersected, a fluid particle at that spatial node would have two distinct velocity directions, violating steady-state constraints.
### Step 1: Final Conclusion
Statements C, D, and E are correct, leading to option (2).
### Pattern Recognition
Contrast Liquid vs Gas viscosity trends under temperature increments: Liquids down (intermolecular bounds weaken), Gases up (random thermal diffusion and collision rates escalate).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q2 (2025)
Consider a completely full cylindrical water tank of height $1.6\mathrm{~m}$ and cross-sectional area $0.5\mathrm{~m}^2$ . It has a small hole in its side at a height $90\mathrm{~cm}$ from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load $50\mathrm{~kg}$ is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is: $(\mathrm{g} = 10\mathrm{~m/s}^2)$
- $3\mathrm{~m/s}$
- $5\mathrm{~m/s}$
- $2\mathrm{~m/s}$
- $4\mathrm{~m/s}$
### Related Formula
Bernoulli's Principle between the top surface (1) and the hole (2):
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$
### Core Logic
Let the top surface be point 1, and the opening of the hole be point 2.
- Height of tank, $H = 1.6\mathrm{~m}$ (thus height of point 1, $h_1 = 1.6\mathrm{~m}$)
- Height of hole, $h_2 = 90\mathrm{~cm} = 0.9\mathrm{~m}$ from the bottom.
- Depth of the hole below top surface, $h = H - h_2 = 1.6 - 0.9 = 0.7\mathrm{~m}$.
- Since area of hole is negligibly small compared to tank area, the velocity of fluid surface at the top is negligible ($v_1 \approx 0$).
- Pressure at point 1 (top surface):
$$P_1 = P_0 + \frac{Mg}{A}$$
where $M = 50\mathrm{~kg}$, $A = 0.5\mathrm{~m}^2$, $P_0$ is atmospheric pressure.
- Pressure at point 2 (outside the hole):
$P_2 = P_0$
### Step 1: Applying Bernoulli's Equation
Substitute these values into Bernoulli's equation with reference datum at the hole ($h_2 = 0$):
$$P_1 + \rho g h = P_2 + \frac{1}{2}\rho v_2^2$$
$$\left(P_0 + \frac{Mg}{A}\right) + \rho g h = P_0 + \frac{1}{2}\rho v_2^2$$
$$\frac{Mg}{A} + \rho g h = \frac{1}{2}\rho v_2^2$$
### Step 2: Substitution and Calculation
Substitute $M = 50\mathrm{~kg}$, $g = 10\mathrm{~m/s}^2$, $A = 0.5\mathrm{~m}^2$, $\rho = 10^3\mathrm{~kg/m}^3$, and $h = 0.7\mathrm{~m}$:
$$\frac{50 \times 10}{0.5} + 10^3 \times 10 \times 0.7 = \frac{1}{2} \times 10^3 \times v_2^2$$
$$1000 + 7000 = 500 \times v_2^2$$
$$8000 = 500 \times v_2^2$$
$$v_2^2 = 16 \implies v_2 = 4\mathrm{~m/s}$$
{{SOL_IMG1}}
### Pattern Recognition
Modified Torricelli's formula with top pressure overload:
$$v = \sqrt{2gh + \frac{2\Delta P}{\rho}}$$
Substitute $\Delta P = \frac{Mg}{A} = \frac{500}{0.5} = 1000\mathrm{~Pa}$. This quickly yields:
$$v = \sqrt{2(10)(0.7) + \frac{2(1000)}{1000}} = \sqrt{14 + 2} = 4\mathrm{~m/s}$$. Keep this shortcut in mind for pressurized tanks!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q3 (2025)
Two liquids A and B have $\theta_{\mathrm{A}}$ and $\theta_{\mathrm{B}}$ as contact angles in a capillary tube. If $K = \cos\theta_{\mathrm{A}} / \cos\theta_{\mathrm{B}}$, then identify the correct statement:
- $K$ is negative, then liquid A and liquid B have convex meniscus.
- $K$ is negative, then liquid A and liquid B have concave meniscus.
- $K$ is negative, then liquid A has concave meniscus and liquid B has convex meniscus.
- $K$ is zero, then liquid A has convex meniscus and liquid B has concave meniscus.
### Related Formula
$$K = \frac{\cos\theta_{\mathrm{A}}}{\cos\theta_{\mathrm{B}}}$$
* Concave meniscus: $\theta < 90^\circ implies \cos\theta > 0$
* Convex meniscus: $\theta > 90^\circ implies \cos\theta < 0$
### Core Logic
For $K$ to be negative, $\cos\theta_{\mathrm{A}}$ and $\cos\theta_{\mathrm{B}}$ must have opposite mathematical signs.
Looking at option (3):
* Liquid A has a concave meniscus $implies \theta_{\mathrm{A}} < 90^\circ implies \cos\theta_{\mathrm{A}} > 0$ (positive).
* Liquid B has a convex meniscus $implies \theta_{\mathrm{B}} > 90^\circ implies \cos\theta_{\mathrm{B}} < 0$ (negative).
This makes the ratio $K = \frac{\text{positive}}{\text{negative}} < 0$ (negative), which accurately validates the statement.
### Pattern Recognition
Meniscus geometry directly correlates with the sign of $\cos\theta$: acute contact angle = concave (wetting fluid), obtuse contact angle = convex (non-wetting fluid).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
