Blog Log in Get Started Free

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

Q: Are these Mechanical Properties of Fluids questions free to attempt?

Yes. You can attempt every Mechanical Properties of Fluids question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Q: Do I need an account to practise Mechanical Properties of Fluids MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Q: Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Practice free Mechanical Properties of Fluids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

A large tank of cross-section area A is filled with water to a height H. A small hole of area 'a' is made at the base of the tank. It takes time $T_{1}$ to decrease the height of water to $\frac{H}{η} (η > 1)$ ; and it takes $T_{2}$ time to take out the rest of water. If $T_{1} = T_{2}$ , then the value of $η$ is

$2 \sqrt{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}] Now, T_{1} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{\frac{H}{η}}] and T_{2} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{\frac{H}{η}} - 0]$

According to problem

$T_{1} = T_{2} ∴ \sqrt{H} - \sqrt{\frac{H}{η}} = \sqrt{\frac{H}{η}} - 0 \Rightarrow \sqrt{H} = 2 \sqrt{\frac{H}{η}} \Rightarrow η = 4$

● ● ● ● ●

As the temperature of water increases, its viscosity

Remains unchanged

Decreases

Increases

Increases or decreases depending on the external pressure

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b)

As the temperature of a liquid increases, the energy of its molecules increases which increases the movement of molecules.so the liquid becomes more fluid. thus viscosity of liquid decreases.

● ● ● ● ●

A small drop of water falls from rest through a large height h in air; the final velocity is

$\propto \sqrt{h}$

$\propto h$

$\propto (1 / h)$

Almost independent of h

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d)

$T e r m i n a l v e l o c i t y v = \frac{2 r^{2} g (σ - ρ)}{9 η}$

● ● ● ● ●

The rate of flow of liquid in a tube of radius r, length l, whose ends are maintained at a pressure difference P is $V = \frac{{πQPr}^{4}}{ηl}$ where $η$ is coefficient of the viscosity and Q is

$\frac{1}{8}$

$\frac{1}{16}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) In this formula, Q=1/8.

● ● ● ● ●

Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes

$4 Q$

$Q$

$\frac{Q}{4}$

$\frac{Q}{8}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d)

$V = \frac{π (∆ P) r^{4}}{8 η l} ∴ V \propto (∆ P) r^{4} (η and l are constants) ∴ \frac{V_{2}}{V_{1}} = (\frac{∆ P_{2}}{∆ P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} ∴ V_{2} = \frac{Q}{8}$

● ● ● ● ●

Water is flowing in a pipe of diameter 4 cm with a velocity 3 m/s. The water then enters into a tube of diameter 2 cm. The velocity of water in the other pipe is

3 m/s

6 m/s

12 m/s

8 m/s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c)

$a_{1} v_{1} = a_{2} v_{2} \Rightarrow \frac{v_{2}}{v_{1}} = \frac{a_{1}}{a_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \Rightarrow v_{2} = 3 \times (2)^{2} = 12 m / s$

● ● ● ● ●

What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are $ρ$ and $σ$ respectively, and the viscosity of the liquid is $η$ ).

$\frac{r^{2} g}{9 η} (ρ - 2 σ)$

$\frac{r^{2} g}{9 η} (2 ρ - σ)$

$\frac{r^{2} g}{9 η} (ρ - σ)$

$\frac{2 r^{2} g}{9 η} (ρ - σ)$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c)

$W h e n t h e a c c e l e r a t i o n o f t h e b a l l i s \frac{g}{2}; F o r c e e q u a t i o n : M g - B - F = M a V ρ g - V σ g - 6 π η r v = V ρ \frac{g}{2} \Rightarrow 6 π η r v = \frac{V ρ g}{2} - V σ g \Rightarrow 6 π η r v = \frac{V g}{2} (ρ - 2 σ) = \frac{4}{3} {πr}^{3} \frac{g}{2} (ρ - 2 σ) \Rightarrow v = \frac{r^{2} g}{9 η} (ρ - 2 σ)$

● ● ● ● ●

Consider the following equation of Bernouilli’s theorem.

$P + \frac{1}{2} {ρV}^{2} + ρgh = K (constant)$

The dimensions of K/P are the same as that of which of the following

Thrust

Pressure

Angle

Viscosity

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$P + \frac{1}{2} {ρV}^{2} + ρgh = K (constant)$

According to principle of homogenecity dimension of K=dimension of $ρ$

Therefore, $\frac{K}{P}$ is dimensionless

out of given options only angle is dimensionless.

So, the dimension of $\frac{K}{P}$ are the same as that of angle.

● ● ● ● ●

An incompressible fluid flows steadily through a cylindrical pipe which has radius 2r at point A and radius r at B further along the flow direction. If the velocity at point A is v, its velocity at point B is

v/2

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$A r e a \times v e l o c i t y = co n s \tan t {πr}^{2} \times v = constant i . e ., {r_{1}}^{2} v_{1} = {r_{2}}^{2} v_{2} given, r_{1} = 2 r_{1} r_{2} = r and v_{1} = v Hence, {(2 r)}^{2} \times v = r^{2} . v_{2} \Rightarrow v_{2} = 4 v$

● ● ● ● ●

A block of ice floats on a liquid of density 1.2 $g / c m^{3}$ in a beaker. Then level of liquid when ice completely melts-

Remains same

Rises

Lowers

(a), (b) or (c)

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) The volume of liquid displaced by floating ice $V_{D} = \frac{M}{σ_{L}}$

Volume of water formed by melting ice, $V_{F} = \frac{M}{σ_{W}}$

If $σ_{L} > σ_{W}$ , then, $\frac{M}{σ_{L}} < \frac{M}{σ_{W}} i . e . V_{D} < V_{F}$

i.e. volume of liquid displaced by floating ice will be lesser than water formed and so the level if liquid will rise.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Mechanical Properties of Fluids question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.