Mechanical Properties of Fluids NEET PYQs (2005–2026) with Solutions

10 solved Mechanical Properties of Fluids PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumBernoulli's principle and equation of continuity

   Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15\ \mathrm{Nm^{-2}}$. The area of cross-section at $P$ and $Q$ are $40\ \mathrm{cm^2}$ and $20\ \mathrm{cm^2}$, respectively. The rate of flow of water through the pipe, in $\mathrm{cm^3s^{-1}}$, is: [Take density of water = $1000\ \mathrm{kg\ m^{-3}}$]

   1. AOption A: $100$
   2. BOption B: $200$
   3. COption C: $300$
   4. DOption D: $400$

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   Correct answer: (D) $400$

   Explanation

   Continuity: $A_P v_P = A_Q v_Q \Rightarrow v_Q = \frac{40}{20}v_P = 2v_P$. Bernoulli for a horizontal pipe: $P_P - P_Q = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{1}{2}\rho(3v_P^2)$. $15 = \frac{1}{2}(1000)(3v_P^2) \Rightarrow v_P^2 = 0.01 \Rightarrow v_P = 0.1\ \mathrm{ms^{-1}} = 10\ \mathrm{cm\ s^{-1}}$. Rate of flow $= A_P v_P = 40\ \mathrm{cm^2} \times 10\ \mathrm{cm\ s^{-1}} = 400\ \mathrm{cm^3s^{-1}}$, option (D).

2. Question 2 (NEET 2026)

   NEET 2026MediumStokes' law and terminal velocity

   In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ($v$) with the ratio of density of spherical ball ($\sigma$) to density of the liquid ($\rho$), is best represented by:

   1. AOption A: Graph A
   2. BOption B: Graph B
   3. COption C: Graph C
   4. DOption D: Graph D

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   Correct answer: (A) Graph A

   Explanation

   The graphs ($v$ against $\sigma/\rho$, with a vertical line at $\sigma/\rho = 1$): (A) $v = 0$ up to $\sigma/\rho = 1$, then a straight line rising from $\sigma/\rho = 1$; (B) a straight line whose dashed extension passes through the origin, so $v \neq 0$ at $\sigma/\rho = 1$; (C) a straight line whose dashed extension meets the axis between 0 and 1, so $v \neq 0$ at $\sigma/\rho = 1$; (D) constant $v$ for $\sigma/\rho > 1$. Stokes' law gives the terminal velocity $v = \frac{2r^2(\sigma - \rho)g}{9\eta} = \frac{2r^2\rho g}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$. For fixed $r$, $\rho$ and $\eta$, $v$ is a linear function of $\frac{\sigma}{\rho}$ that is zero at $\frac{\sigma}{\rho} = 1$ (the ball neither sinks nor rises) and increases linearly for $\frac{\sigma}{\rho} > 1$. This is the graph in option (A).

3. Question 3 (NEET 2026)

   NEET 2026EasyPressure & Pascal's Law

   A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (Consider density of water = 1000 kg m⁻³, 1 atm = 1 × 10⁵ Pa and gravitational acceleration g = 10 m/s²)

   1. AOption A: 990 m
   2. BOption B: 9900 m
   3. COption C: 99 m
   4. DOption D: 9000 m

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   Correct answer: (A) 990 m

   Explanation

   Absolute pressure: P=P₀+ρgh. Therefore 100×10⁵=10⁵+1000×10×h ⇒ h=990 m. Hence option (A) is correct.

4. Question 4 (NEET 2025)

   NEET 2025HardSurface tension, angle of contact, capillary rise

   Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ₀ (θ₀ << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is : (take θ(x) = sin θ(x) = tan θ(x) = dy/dx, g is the acceleration due to gravity)

   1. AOption A: d²y/dx² = (ρg/S)x
   2. BOption B: d²y/dx² = (ρg/S)y
   3. COption C: d²y/dx² = √(ρg/S)
   4. DOption D: dy/dx = √(ρg/S)x

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   Correct answer: (B) d²y/dx² = (ρg/S)y

   Explanation

   For a curved liquid surface in equilibrium, the excess pressure due to surface tension balances the hydrostatic pressure. Using small angle approximation, curvature is given by d²y/dx² and balancing forces gives: d²y/dx² = (ρg/S)y.

5. Question 5 (NEET 2025)

   NEET 2025HardSurface Tension & Capillarity

   A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$. It is filled with a gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to $0$ (zero) in time $T$. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^a$ and $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$, then

   1. AOption A: $a=\frac{1}{2},\ \alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=1,\ \delta=\frac{3}{2}$
   2. BOption B: $a=-\frac{1}{2},\ \alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=-\frac{1}{2},\ \delta=\frac{5}{2}$
   3. COption C: $a=-\frac{1}{2},\ \alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=\frac{1}{2},\ \delta=\frac{7}{2}$
   4. DOption D: $a=\frac{1}{2},\ \alpha=\frac{1}{2},\ \beta=-\frac{1}{2},\ \gamma=\frac{1}{2},\ \delta=\frac{7}{2}$

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   Correct answer: (C) $a=-\frac{1}{2},\ \alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=\frac{1}{2},\ \delta=\frac{7}{2}$

   Explanation

   Pressure inside the balloon due to surface tension is proportional to: $$\Delta P \propto \frac{S}{r}$$ Using Bernoulli’s principle: $$\frac{1}{2}\rho v^2 \propto \frac{S}{r}$$ Hence, $$v(r) \propto \sqrt{\frac{S}{\rho r}} \propto r^{-1/2}$$ Therefore, $$a=-\frac{1}{2}$$ Rate equation: $$A v(r)\, dt \propto r^2 dr$$ Substituting $v(r)$:$$dt \propto \frac{r^2 dr}{A\sqrt{S/(\rho r)}} = \frac{\rho^{1/2}}{A S^{1/2}} r^{5/2} dr$$Integrating from$0$to$R$:$$T \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$$Thus,$$\alpha=-\frac{1}{2},\ \beta=-1,\ \gamma=\frac{1}{2},\ \delta=\frac{7}{2}$$ Hence option (C) is correct.

6. Question 6 (NEET 2024)

   NEET 2024MediumSurface Tension and Excess Force

   A thin flat circular disc of radius $4.5\,\text{cm}$ is placed gently over the surface of water. If surface tension of water is $0.07\,\text{N m}^{-1}$, then the excess force required to take it away from the surface is:

   1. AOption A: $1.98\,\text{mN}$
   2. BOption B: $99\,\text{N}$
   3. COption C: $19.8\,\text{mN}$
   4. DOption D: $198\,\text{N}$

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   Correct answer: (C) $19.8\,\text{mN}$

   Explanation

   The excess force required to detach the disc from the water surface is: $$F = 2 \times (2\pi r)T$$ where: $$r = 4.5\,\text{cm} = 0.045\,\text{m}$$ $$T = 0.07\,\text{N m}^{-1}$$ Substituting: $$F = 4\pi (0.045)(0.07)$$ $$F \approx 0.0396\,\text{N}$$ Considering one liquid-air interface for the thin disc: $$F = 2\pi rT$$ $$F = 2\pi(0.045)(0.07)$$ $$F \approx 0.0198\,\text{N}$$ $$F = 19.8\,\text{mN}$$ Hence, the correct answer is: **(3) $19.8\,\text{mN}$**.

7. Question 7 (NEET 2023)

   NEET 2023MediumSurface Tension & Capillarity

   The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly : (surface tension of soap solution = 0.03 $N m^−1$)

   1. AOption A: 30.16 × 10⁻⁴ J
   2. BOption B: 5.06 × 10⁻⁴ J
   3. COption C: 3.01 × 10⁻⁴ J
   4. DOption D: 50.1 × 10⁻⁴ J

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   Correct answer: (C) 3.01 × 10⁻⁴ J

   Explanation

   Energy required to form a soap bubble is equal to the surface energy: E = 2 × 4πr²T where r = 2 cm = 0.02 m and T = 0.03 N/m. E = 8π(0.02)²(0.03) ≈ 3.01×10−4 J Hence, option (C) is correct.

8. Question 8 (NEET 2023)

   NEET 2023EasyBernoulli's Principle

   The venturi-meter works on :

   1. AOption A: Huygen’s principle
   2. BOption B: Bernoulli’s principle
   3. COption C: The principle of parallel axes
   4. DOption D: The principle of perpendicular axes

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   Correct answer: (B) Bernoulli’s principle

   Explanation

   Venturi-meter measures fluid flow speed using pressure differences produced due to varying cross-sectional areas. It works on Bernoulli’s principle. Therefore, option (B) is correct.

9. Question 9 (NEET 2022)

   NEET 2022EasySurface Tension & Capillarity

   If a soap bubble expands, the pressure inside the bubble:

   1. AOption A: Is equal to the atmospheric pressure
   2. BOption B: Decreases
   3. COption C: Increases
   4. DOption D: Remains the same

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   Correct answer: (B) Decreases

   Explanation

   Excess pressure inside a soap bubble is: ΔP = 4T/r As the radius increases during expansion, excess pressure decreases. Hence pressure inside the bubble decreases. Therefore option (B) is correct.

10. Question 10 (NEET 2022)

    NEET 2022MediumViscosity & Stokes' Law

    A spherical ball is dropped in a long column of highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t), is:

    1. AOption A: D
    2. BOption B: A
    3. COption C: B
    4. DOption D: C

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    Correct answer: (C) B

    Explanation

    A ball falling through viscous liquid attains terminal velocity. Velocity increases rapidly initially and gradually approaches a constant value. Curve B represents this behaviour. Hence option (C) is correct.