NEET Mechanical Properties of Fluids Previous Year Questions with Solutions

Because the tank is very wide in the z-direction, the liquid surface is cylindrical (curvature in z-direction is zero). The Laplace excess pressure across a surface with radius of curvature R is governed by surface tension S.

• • For a surface profile y(x) with small slope (dy/dx ≪ 1), the curvature simplifies to 1/R ≈ d²y/dx².

At any horizontal position x, the atmospheric pressure acts above the surface. At the flat baseline level (y = 0), the pressure is atmospheric. The hydrostatic pressure difference between the liquid at height y(x) and the flat surface is ρgy.

• • Equilibrium requires excess Laplace pressure to balance the gravitational head.

Equating the Laplace pressure to the hydrostatic pressure head yields the second-order linear differential equation governing the meniscus profile.

• • This is the classical capillary rise meniscus differential equation with characteristic capillary length λ_c = √(S/ρg).

The excess pressure inside a spherical elastic balloon of instantaneous radius r with effective surface tension S is given by Laplace’s law:

• • Applying Bernoulli’s theorem between the high-pressure gas inside the balloon and the atmospheric exhaust outside through the small opening of area A:
• • ΔP = (1/2) ρ v² ⟹ v(r) = √(2ΔP / ρ) ∝ √(S / (ρ r)) = (S / ρ)^(1/2) r^(-1/2).
• • Comparing with v(r) ∝ r^a directly gives: a = -1/2.

By conservation of mass / volume continuity, the rate of decrease of the balloon volume equals the volume flux exiting through the aperture:

• • Rearranging to separate variables r and t:
• • -r^2 \cdot r^{1/2} dr = -r^{5/2} dr = \left(\frac{A}{4\pi}\sqrt{\frac{8S}{\rho}}\right) dt.

Integrate from the initial radius r = R at t = 0 to r = 0 at t = T:

• • Solving for total deflation time T:
• • T ∝ S^(-1/2) A^(-1) ρ^(1/2) R^(7/2).
• • Comparing with T ∝ S^α A^β ρ^γ R^δ:
• • α = -1/2, β = -1, γ = 1/2, δ = 7/2.
• • Along with a = -1/2, this corresponds uniquely to Option (3).

When a flat solid circular disc of radius R rests on the surface of water, the water film clings to the boundary circumference of the disc. The perimeter along which the surface tension acts downwards as the disc is pulled upward is:

Surface tension T is force per unit length. The additional upward force F needed to overcome surface tension and detach the disc is:

Substitute T = 0.07 N m⁻¹ and R = 4.5 cm = 0.045 m:

Using π ≈ 3.1416:

A Venturi-meter is a device used for measuring the speed of flow of an incompressible fluid in a pipe. It consists of a broad entrance tube of area A₁, a constricted throat of area A₂, and a manometer to record the pressure difference between the two sections.

By the equation of continuity, as area decreases at the throat (A₂ < A₁), the fluid velocity increases (v₂ > v₁):

For horizontal streamline flow, Bernoulli's principle states that the total mechanical energy per unit volume is constant:

Thus, the functioning of a Venturi-meter is fundamentally based on Bernoulli's principle.

Unlike a solid liquid drop which has only one outer surface, a soap bubble has a hollow spherical air pocket inside and air outside, creating two distinct liquid-air interfaces of virtually identical surface area:

The work done W required to form the bubble from a soap solution is equal to the surface energy created:

Substitute radius r = 2 × 10⁻² m and surface tension T = 0.03 N/m:

Using π ≈ 3.1416:

When the small sphere attains a constant terminal velocity ($v_t$), its downward acceleration is zero ($a = 0$). By Newton's second law, the downward gravitational force is perfectly balanced by the sum of upward buoyant force ($F_B$) and upward viscous drag ($F_v$):

Let $V$ be the volume of the sphere. The mass is $M = V \cdot d$, so $V = M / d$. The weight is $W = Mg$. The buoyant force exerted by glycerine of density $\rho_L = d/2$ is:

Substitute $W = Mg$ and $F_B = Mg/2$ into the force balance equation:

When a capillary tube of internal radius $$r$$ is immersed in water, the meniscus forms a contact angle $$\theta$$. Surface tension $$T$$ acts along the contact perimeter $$2\pi r$$. The upward vertical force supporting the liquid column is: $$F_{\text{up}} = 2\pi r T \cos\theta$$.

At equilibrium, this upward force balances the weight of the liquid inside the capillary: $$W = m g$$ $$m g = 2\pi r T \cos\theta \implies m = \frac{2\pi r T \cos\theta}{g}$$.

Since surface tension $$T$$, contact angle $$\theta$$, and $$g$$ are constant for the water-glass system: $$m \propto r$$ (Note: Even though height $$h = \frac{2T\cos\theta}{\rho g r} \propto \frac{1}{r}$$ halves, the cross-sectional area $$A = \pi r^2$$ quadruples, so volume $$V = A h \propto r^2 \times (1/r) = r$$, confirming $$m \propto r$$).

For the second tube with radius $$r_2 = 2r$$: $$m_2 = 2 \times m_1 = 2 \times 5\text{ g} = 10.0\text{ g}$$. Matches Option 3.