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

● ● ● ● ●

If the radius of a soap bubble is four times that of another, then the ratio of their pressures will be

1 : 4

4 : 1

16 : 1

1 : 16

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

Explanation

(a) $∆ P \propto \frac{1}{r} \Rightarrow \frac{∆ P_{1}}{∆ P_{2}} = \frac{r_{2}}{r_{1}} = \frac{r}{4 r} = \frac{1}{4}$

● ● ● ● ●

A spherical drop of water has radius 1 mm If surface tension of water is $70 \times 10^{- 3}$ N/m difference of pressures between inside and out side of the spherical drop is

$35 N / m^{2}$

$70 N / m^{2}$

$140 N / m^{2}$

Zero

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

Explanation

● ● ● ● ●

In capillary tube, pressure below the curved surface of water will be

Equal to atmospheric

Equal to upper side pressure

More than upper side pressure

Lesser than upper side pressure

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

Explanation

(d)

● ● ● ● ●

The pressure inside a small air bubble of radius 0.1 mm situated just below the surface of water will be equal to
[Take surface tension of water $70 \times 10^{- 3} {Nm}^{- 1}$ and atmospheric pressure = $1.013 \times 10^{5} {Nm}^{- 2}$ ]

$2.054 \times 10^{3} Pa$

$1.027 \times 10^{3} Pa$

$1.027 \times 10^{5} Pa$

$2.054 \times 10^{5} Pa$

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

Explanation

$\Rightarrow P_{in} - P_{out} = \frac{2 T}{r} = \frac{2 \times 70 \times 10^{- 3}}{0.1 \times 10^{- 3}} = 1400 Pa \Rightarrow P_{in} = 1400 + 1.013 \times 10^{5} = 1.027 \times 10^{5} Pa$

● ● ● ● ●

Two bubbles A and B are joined through a narrow tube where bubble A is bigger. Then

The size of A will increase

The size of B will increase

The size of B will increase until the pressure equals

None of these

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

Explanation

(a)

$r_{A} > r_{B} and P \propto \frac{1}{r} so P_{A} < P_{B}$

So air will flow from B to A i.e. size of A will increase.

● ● ● ● ●

Two soap bubbles have different radii but their surface tension is the same. Mark the correct statement

Internal pressure of the smaller bubble is higher than the internal pressure of the larger bubble

Pressure of the larger bubble is higher than the smaller bubble

Both bubbles have the same internal pressure

None of the above

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

Explanation

(a) $∆ P = \frac{4 T}{R} ∴ ∆ P \propto \frac{1}{R}$ (T= constant)

Hence, the internal pressure of smaller bubble is larger than that of larger bubble.

● ● ● ● ●

If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm, then the surface tension of soap solution will be (r = 1 cm and density d = 0.8 gm/cc)

3.9 N/m

$3.9 \times 10^{- 2} N / m$

$3.9 \times 10^{- 3} N / m$

3.9 dyne/m

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

Explanation

(b) $\frac{4 T}{R} = hdg ∴ T = \frac{Rhdg}{4} T = \frac{10^{- 2} \times 2 \times 10^{- 3} \times 0.8 \times 10^{3} \times 9.8}{4} = 3.9 \times 10^{- 2} N / m$

● ● ● ● ●

Due to capillary action, a liquid will rise in a tube, if the angle of contact is

Acute

Obtuse

90°

Zero

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

Explanation

(a) $h = \frac{2 Tcosθ}{rdg}$ . If $θ$ is less than 90° then h will be positive

● ● ● ● ●

Two parallel glass plates are dipped partly in the liquid of density 'd' keeping them vertical. If the distance between the plates is 'x', surface tension for liquids is T and angle of contact is $θ$ , then rise of liquid between the plates due to capillary will be

$\frac{T cosθ}{xd}$

$\frac{2 T cosθ}{xdg}$

$\frac{2 T}{xdgcosθ}$

$\frac{T cosθ}{xdg}$

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

Explanation

The rise of liquid between two parallel glass plates due to capillary action is given by the formula (2T cosÎ¸)/(xdg), where T is the surface tension, Î¸ is the angle of contact, x is the distance between the plates, d is the density of the liquid, and g is the acceleration due to gravity.

● ● ● ● ●

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{- 4} N$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{- 2} {Nm}^{- 1}$ , the inner circumference of the capillary must be

$1.25 \times 10^{- 2} m$

$0.50 \times 10^{- 2} m$

$6.5 \times 10^{- 2} m$

$12.5 \times 10^{- 2} m$

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

Explanation

(d)

$6 \times 10^{- 2} \times Circumference = Force ∴ Circumference = \frac{75 \times 10^{- 4}}{6 \times 10^{- 2}} = 12.5 \times 10^{- 2} m$

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.