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Mechanical Properties of Fluids MCQs for NEET — Physics Questions with Answers

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We have two (narrow) capillary tubes T₁ and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

4 cm³/sec

(16/3) cm³/sec

(8/17) cm³/sec

None of these

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Explanation

(b) $V = \frac{{Pπr}^{4}}{8 ηl} \Rightarrow P = \frac{V 8 ηl}{{πr}^{4}}$

For composite tube

$V_{1} = \frac{{Pπr}^{4}}{8 η (l + \frac{l}{2})} = \frac{2}{3} \frac{{πPr}^{4}}{8 ηl} = \frac{2}{3} \times 8 = \frac{16}{3} \frac{{cm}^{3}}{\sec} [∵ l_{1} = l = 2 l_{2} or l_{2} = \frac{l}{2}]$

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The Reynolds number of a flow is the ratio of

Gravity to viscous force

Gravity force to pressure force

Inertia forces to viscous force

Viscous forces to pressure forces

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Explanation

(c) $R_{e} = \frac{e v d}{η} o r R_{e} = \frac{e v^{2} d \times A}{A \times v \times η} \Rightarrow R_{e} = \frac{e v^{2} A}{η A v / d}$

Here, $η A v / d$ =viscous force and $e v^{2} A =$ Inertial force

Hence, $R_{e} = \frac{I n e r t i a l f o r c e}{v i s c o u s f o r c e}$

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Water is flowing through a tube of non-uniform cross-section ratio of the radius at entry and exit end of the pipe is 3 : 2. Then the ratio of velocities at entry and exit of liquid is -

4 : 9

9 : 4

8 : 27

1 : 1

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Explanation

(a) If velocities of water at entry and exit points are v₁ and v₂, then according to equation of continuity,

$A_{1} V_{1} = A_{2} V_{2} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = {(\frac{r_{2}}{r_{1}})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$

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An application of Bernoulli's equation for fluid flow is found in

Dynamic lift of an aeroplane

Viscosity meter

Capillary rise

Hydraulic press

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Explanation

Bernoulli's equation relates the pressure, velocity, and height of a flowing fluid. It is used to explain the dynamic lift of an aeroplane, where the faster air flow over the curved upper surface creates a lower pressure, resulting in an upward force on the wing.

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The Working of an atomizer depends upon

Bernoulli's theorem

Boyle's law

Archimedes principle

Newton's law of motion

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Explanation

(a)

Atomizer works on the principle of liquid flow, It follows bernaulis principle because here, horizontal air passing causes the vertices liquid to come out.

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The pans of a physical balance are in equilibrium. Air is blown under the right hand pan; then the right hand pan will

Move up

Move down

Move erratically

Remain at the same level

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Explanation

(b) According to Bernoulli's theorem.

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According to Bernoulli's equation

$\frac{P}{ρg} + h + \frac{1}{2} \frac{v^{2}}{g} = constant$

The terms A, B and C are generally called respectively:

Gravitational head, pressure head and velocity head

Gravity, gravitational head and velocity head

Pressure head, gravitational head and velocity head

Gravity, pressure and velocity head

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Explanation

(c)

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A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 ${kgm}^{- 3}$ . The velocity with which gasoline begins to shoot out of the hole is

27.8 ${ms}^{- 1}$

41.0 ${ms}^{- 1}$

9.6 ${ms}^{- 1}$

19.7 ${ms}^{- 1}$

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Explanation

The velocity of the gasoline shooting out of the hole can be calculated using Bernoulli's equation. The pressure difference between the inside of the tank and the atmospheric pressure provides the driving force for the flow, and the velocity depends on the height of the gasoline column.

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A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water

9 minute

7 minute

5 minute

3 minute

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Explanation

(b) Time taken to be emptied for h height, $t = \sqrt{\frac{2 h}{g}}$

and for $\frac{h}{2}$ height , $t = \sqrt{\frac{2 h / 2}{g}} = \sqrt{\frac{h}{g}} ∴ \frac{t'}{t} = \frac{1}{\sqrt{2}} \Rightarrow t' = \frac{t}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 7 minute$

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A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{2 h}{g} \cdot \frac{D}{d}}$

$\sqrt{\frac{2 h}{g} \cdot \frac{d}{D}}$

$\sqrt{\frac{2 h}{g}} (\frac{d}{D - d})$

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Explanation

(d) Upthrust – weight of body = apparent weight

VDg-Vdg = Vda,

Where a = retardation of body $∴ a = (\frac{D - d}{d}) g$

The velocity gained after fall from h height in air,

$v = \sqrt{2 gh}$

Hence, time to come in rest,

$t = \frac{v}{a} = \frac{\sqrt{2 gh} \times d}{(D - d) g} = \sqrt{\frac{2 h}{g}} \times \frac{d}{(D - d)}$

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