Mechanical Properties of Fluids Previous Year Questions — JEE Main Physics
32 past-year Mechanical Properties of Fluids questions from JEE Main (Physics) — page 4.
Q58 (2024)
A plane is in level flight at constant speed and each of its two wings has an area of $40\mathrm{~m}^2$. If the speed of the air is $180\mathrm{~km/h}$ over the lower wing surface and $252\mathrm{~km/h}$ over the upper wing surface, the mass of the plane is _______ $\mathrm{kg}$. (Take air density to be $1\mathrm{~kg\ m}^{-3}$ and $g = 10\mathrm{~ms}^{-2}$)
### Related Formula
Bernoulli's pressure balance equation for aerofoils:
$$\Delta P = P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$
Dynamic Lift force balancing plane weight:
$$F_{\text{lift}} = \Delta P \cdot A_{\text{total}} = mg$$
### Core Logic
Convert velocity limits to SI units:
$$v_1 = 180\mathrm{~km/h} = 180 \times \frac{5}{18} = 50\mathrm{~ms}^{-1}$$
$$v_2 = 252\mathrm{~km/h} = 252 \times \frac{5}{18} = 70\mathrm{~ms}^{-1}$$
Total effective wing area layout (2 wings):
$$A_{\text{total}} = 2 \times 40 = 80\mathrm{~m}^2$$
### Step 1: Calculate Mass Balance
Substitute these values into the dynamic lift equation:
$$mg = \frac{1}{2} \rho (v_2^2 - v_1^2) A_{\text{total}}$$
$$m(10) = \frac{1}{2} \times 1 \times (70^2 - 50^2) \times 80$$
$$10m = 40 \times (4900 - 2500) = 40 \times 2400 = 96000$$
$$m = 9600\mathrm{~kg}$$
### Pattern Recognition
Remember to multiply individual wing areas by 2 for standard multi-wing lift structures ($A_{\text{total}} = 2A$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q1451 (2024)
A small liquid drop of radius $R$ is divided into $27$ identical liquid drops. If the surface tension is $T$, then the work done in the process will be:
- $8\pi R^2 T$
- $3\pi R^2 T$
- $\frac{1}{8}\pi R^2 T$
- $4\pi R^2 T$
### Related Formula
The work done in changing the surface area of a liquid is given by:
$$W = T \cdot \Delta A$$
where:
* $T$ is the surface tension of the liquid.
* $\Delta A = A_f - A_i$ is the change in the total surface area.
### Core Logic
Since the total volume remains constant during splitting:
$V_i = V_f$
$$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$$
$$R^3 = 27r^3 \implies r = \frac{R}{3}$$
### Step 1: Calculate the Change in Surface Area
Initial surface area of the single drop:
$A_i = 4\pi R^2$
Final surface area of $27$ small drops:
$$A_f = 27 \times (4\pi r^2) = 27 \times 4\pi \left(\frac{R}{3}\right)^2$$
$$A_f = 27 \times 4\pi \frac{R^2}{9} = 12\pi R^2$$
Change in surface area:
$$\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$
### Step 2: Calculate Work Done
Substituting the change in area into the work done formula:
$$W = T \cdot \Delta A = 8\pi R^2 T$$
### Pattern Recognition
For splitting a large drop of radius $R$ into $n$ identical small drops, the change in surface area is given by $\Delta A = 4\pi R^2 (n^{1/3} - 1)$. Substituting $n = 27$ gives $\Delta A = 4\pi R^2 (3 - 1) = 8\pi R^2$, leading immediately to $8\pi R^2 T$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q32 (2024)
Given below are two statements:
Statement (I): Viscosity of gases is greater than that of liquids.
Statement (II): Surface tension of a liquid decreases due to the presence of insoluble impurities.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $\text{Statement I is correct but Statement II is incorrect}$
- $\text{Statement I is incorrect but Statement II is correct}$
- $\text{Both Statement I and Statement II are incorrect}$
- $\text{Both Statement I and Statement II are correct}$
### Core Logic
Statement (I): Liquids have much stronger intermolecular forces compared to gases, leading to significantly higher viscosity in liquids than in gases. Thus, Statement I is incorrect.
Statement (II): The presence of insoluble impurities (like soap or detergents) disrupts the cohesive forces between liquid molecules at the surface, which decreases the surface tension. Thus, Statement II is correct.
### Pattern Recognition
Viscosity in liquids decreases with temperature, whereas in gases it increases with temperature due to molecular collisions. Insoluble impurities act as surface-active agents that lower surface tension cohesive stability.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q33 (2024)
Given below are two statements:
Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water.
Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water.
In the light of the above statements, choose the most appropriate from the options given below:
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
### Related Formula
The height of capillary rise ($h$) is given by:
$$h = \frac{2T \cos \theta}{\rho g r}$$
where,
$T$ = surface tension of the liquid
$\theta$ = angle of contact
$\rho$ = density of the liquid
$r$ = radius of the capillary tube
### Core Logic
As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension ($T$).
Since $h \propto T$ (assuming $\rho$ and $\theta$ remain relatively constant), height of capillary rise decreases with increase in temperature.
### Step 1: Evaluate Statement I and II
Because $T_{\text{hot}} \lt T_{\text{cold}}$, it follows that $h_{\text{hot}} \lt h_{\text{cold}}$.
* **Statement I** states: the height of capillary rise will be smaller in hot water. This is **True**.
* **Statement II** states: the height of capillary rise will be smaller in cold water. This is **False**.
### Pattern Recognition
Remember the key physical dependence: **Temperature Up $\implies$ Surface Tension Down $\implies$ Capillary Rise Down**. This basic trend of cohesive/adhesive properties versus thermal energy frequently appears in competitive conceptual physical chemistry/fluid physics questions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q60 (2024)
In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70~\mathrm{ms}^{-1}$ and $65~\mathrm{ms}^{-1}$ respectively. If the wing area is $2~\mathrm{m}^2$ the lift of the wing is ________ N. (Given density of air $= 1.2~\mathrm{kg}\,\mathrm{m}^{-3}$)
### Related Formula
From Bernoulli's Principle (ignoring small height differences across the wing thickness):
$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$
$$\Delta P = P_2 - P_1 = \frac{1}{2} \rho (v_1^2 - v_2^2)$$
Net aerodynamic lift force ($F$) acting on wing area $A$ is:
$$F = \Delta P \cdot A = \frac{1}{2} \rho (v_1^2 - v_2^2) A$$
### Core Logic
Given values:
* Flow speed on upper surface ($v_1$) = $70 \mathrm{~ms}^{-1}$
* Flow speed on lower surface ($v_2$) = $65 \mathrm{~ms}^{-1}$
* Wing area ($A$) = $2 \mathrm{~m}^2$
* Density of air ($\rho$) = $1.2 \mathrm{~kg\cdot m}^{-3}$
### Step 1: Compute Lift Force
Substituting these metrics directly into the dynamic lift equation:
$$F = \frac{1}{2} \times 1.2 \times \left(70^2 - 65^2\right) \times 2$$
Notice that the factors $\frac{1}{2}$ and $2$ cancel out perfectly:
$$F = 1.2 \times \left(4900 - 4225\right)$$
$$F = 1.2 \times 675 = 810 \mathrm{~N}$$
Therefore, the net lift force on the wing is $810 \mathrm{~N}$.
### Pattern Recognition
Use the algebraic difference of squares shortcut $a^2 - b^2 = (a-b)(a+b)$ to solve velocity square differences quickly: $70^2 - 65^2 = (70-65)(70+65) = 5 \times 135 = 675$, bypassing squaring heavy multi-digit calculations.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
