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All Physics Chemistry Botany Zoology

Language English हिंदी

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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

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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.

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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

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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.

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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

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Explanation

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

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

Acute

Obtuse

90°

Zero

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Explanation

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

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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

$\frac{T cosθ}{xd}$

$\frac{2 T cosθ}{xdg}$

$\frac{2 T}{xdgcosθ}$

$\frac{T cosθ}{xdg}$

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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.

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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

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

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

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

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

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Explanation

(d)

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

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Two capillary tubes P and Q are dipped in water. The height of water level in capillary P is 2/3 to the height in Q capillary. The ratio of their diameters is

2 : 3

3 : 2

3 : 4

4 : 3

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Explanation

(b) $r \propto \frac{1}{h} \Rightarrow \frac{r_{p}}{r_{Q}} = \frac{h_{Q}}{h_{P}} = \frac{h}{\frac{2}{3} h} = \frac{3}{2}$

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A vessel, whose bottom has round holes with diameter of 0.1mm, is filled with water. The maximum height to which the water can be filled without leakage is
(S.T. of water =75 dyne/cm, g =1000 $cm / s^{2}$ )

100 cm

75 cm

50 cm

30 cm

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Explanation

(d) $h = \frac{2 T}{rdg} = \frac{2 \times 75}{0.005 \times 1 \times 10^{3}} = 30 cm$

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Water rise in a capillary tube when its one end is dipped vertically in it, is 3 cm. If the surface tension of water is $75 \times 10^{- 3} N / m$ , then the diameter of capillary will be

0.1 mm

0.5 mm

1.0 mm

2.0 mm

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Explanation

(c)

$T = \frac{rhρg}{2} \Rightarrow 75 \times 10^{- 3} = \frac{3 \times 10^{- 2} \times r \times 10^{3} \times 9.8}{2} \Rightarrow r = \frac{1}{2} mm ∴ D = 2 r = 1 mm$

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By inserting a capillary tube upto a depth l in water, the water rises to a height h ( h<l). If the lower end of the capillary is closed inside water and the capillary is taken out and closed end opened, to what height the water will remain in the tube ?

Zero

l+h

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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.