Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

A liquid wets a solid completely. The meniscus of the liquid in a sufficiently long tube is

Flat

Concave

Convex

Cylindrical

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

Explanation

(b) The liquid wets the solid completely, so angle of contact, $θ < 90 °$ . Then the meniscus will be concave when $θ < 90 °$ .

● ● ● ● ●

For which of the two pairs, the angle of contact is same

Water and glass; glass and mercury

Pure water and glass; glass and alcohol

Silver and water; mercury and glass

Silver and chromium; water and chromium

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

Explanation

(b) Both liquids water and alcohol have same nature (i.e. wet the solid). Hence angle of contact for both is acute

● ● ● ● ●

If the surface of a liquid is plane, then the angle of contact of the liquid with the walls of container is

Acute angle

Obtuse angle

90°

0°

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

Explanation

(d) Tangent drawn at point of contact makes 0° with wall of container.

● ● ● ● ●

If two soap bubbles of different radii are in communication with each other

Air flows from larger bubble into the smaller one

The size of the bubbles remains the same

Air flows from the smaller bubble into the large one and the larger bubble grows at the expense of the smaller one

The air flows from the larger

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

Explanation

● ● ● ● ●

The surface tension of soap solution is $25 \times 10^{- 3} {Nm}^{- 1}$ . The excess pressure inside a soap bubble of diameter 1 cm is

10 Pa

20 Pa

5 Pa

None of the above

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

Explanation

(b) Excess pressure $∆ P = \frac{4 T}{r}$

$= \frac{4 \times 2 \times 25 \times 10^{- 3}}{1 \times 10^{- 2}} = 20 N / m^{2} = 20 Pa (as r = d / 2)$

● ● ● ● ●

When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

$r_{2} - r_{1}$

$\frac{r_{2} - r_{1}}{r_{2} r_{1}}$

$\frac{r_{2} r_{1}}{r_{2} - r_{1}}$

$r_{2} + r_{1}$

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

Explanation

$\Rightarrow \frac{4 T}{r} = \frac{4 T}{r_{2}} - \frac{4 T}{r_{1}} \Rightarrow r = \frac{r_{1} r_{2}}{r_{1} - r_{2}}$

● ● ● ● ●

The excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional to

$r^{2}$

$r^{- 1}$

$r^{- 2}$

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

Explanation

● ● ● ● ●

A long cylindrical glass vessel has a small hole of radius 'r' at its bottom. The depth to which the vessel can be lowered vertically in the deep water bath (surface tension T) without any water entering inside is

$4 T / rρg$

$3 T / rρg$

$2 T / rρg$

$T / rρg$

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

Explanation

● ● ● ● ●

The excess of pressure inside a soap bubble than that of the outer pressure is

$\frac{2 T}{r}$

$\frac{4 T}{r}$

$\frac{T}{2 r}$

$\frac{T}{r}$

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

Explanation

(b) The excess of pressure inside a soap bubble = $\frac{4 T}{r}$

● ● ● ● ●

The pressure of air in a soap bubble of 0.7cm diameter is 8 mm of water above the pressure outside. The surface tension of the soap solution is

100 dyne/cm

68.66 dyne/cm

137 dyne/cm

150 dyne/cm

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

Explanation

(b) $∆ P = \frac{4 T}{r} = hdg \Rightarrow T = \frac{rhdg}{4} = \frac{0.35 \times 0.8 \times 1 \times 10^{3}}{4} = 70 dyne / cm \equiv 68.66 dyne / cm$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET 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.