Surface Tension Test

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Surface Tension Test - 8

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Question 1
1 / -0

A soap bubble is blown with the help of a mechanical pump at the mouth of a tube. The pump produces a certain increase per minute in the volume of the bubble, irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time t will be-

A

B

C

D

Solution

ΔP = 4T/r ∴ ΔP ∝ 1/r

As increase in volume is directly proportional to time, radius of soap bubble increases with time.

∴ ΔP ∝ 1/t
Question 2
1 / -0

Which graph represents the variation of surface tension with temperature over small temperature ranges for water?

A

B

C

D

Solution

T_C= T₀(1−αt) i.e. surface tension decreases with increase in temperature.
Question 3
1 / -0

A rectangular film of liquid is extended from (4 cm x 2 cm) to (5 cm x 4 cm). If the work done is 3 x 10^-4 J, the value of the surface tension of the liquid is:

A
0.250 Nm^-1

B
0.125 Nm^-1

C
0.2 Nm^-1

D
8.0 Nm^-1

Solution

Work done=change in surface area x surface tension

Change in area, A=(5x4 - 4x2)x2 ( film has two surfaces)

= (20-8) x 2 cm²=24 cm²

=24 x 10^-4 m²

So, the work done, W = T.A

3x10^-4=Tx24x10^-4

T=0.125 N/m
Question 4
1 / -0

The wettability of a surface by a liquid depends primarily on:

A
surface tension

B
density

C
angle of contact between the surface and the liquid

D
viscosity

Solution

The wettability of a surface by a liquid depends on the angle of contact between the surface and the liquid.
Question 5
1 / -0

A soap bubble, having a radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5×10⁻²N/m. The pressure inside the bubble equals at a point Z₀ below the free surface of the water in a container. Taking g= 10 m/s², density of water = 10³kg/m³, the value of Z₀ is:

A
0.5 cm

B
100 cm

C
10 cm

D
1 cm

Solution

A soap bubble has two surfaces. So excess pressure inside the soap buble = 4S/r
Question 6
1 / -0

The surface tension of the soap solution is 2 x 10⁻²² N/m. If a soap bubble of radii 4 cm is blown, then the amount of work done is:

A
4π x 10⁻⁶j

B
2.56\π x 10^-4 J

C
16π x 10^-5 J10-5 J

D
16π x ¹⁰−6 J

Solution

W=T×ΔS
Question 7
1 / -0

Two water droplets merge with each other to form a larger droplet isothermally. In this process

A
Energy is absorbed

B
Energy is liberated

C
Energy is neither liberated nor absorbed

D
Some mass is converted into energy

Solution

When water droplets merge to form a bigger drop the total surface area decreases. Since the molecules in the surface have greater potential energy, the potential energy of surface molecules in the bigger drop decreases from before the merge.

So, the surface energy per unit area is equal to the surface tension. Since surface tension remains the same, the surface energy will be less in the merged bigger drop.

Therefore, energy is liberated in the process.
Question 8
1 / -0

A liquid does not wet the solid surface if the angle of contact is :

A
equal to 45°

B
equal to 60°

C
greater than 90°

D
zero

Solution

If the angle of contact is greater than 90^o then a liquid does wet the solid surface.
Question 9
1 / -0

A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10^–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. The surface tension of the film is:

A
3.5×10⁻² N m⁻¹

B
2.5×10⁻² N m⁻¹

C
3.5×10⁻³ N m⁻¹

D
2.5 × 10^-3 N m⁻¹

Solution

l=2×30 cm=0.60 ml=2×30 cm=0.60 m

Step 2: Calculate the surface tension by balancing the forces.

Let T be the surface tension.

If F is the total force on the slider due to surface tension,
Question 10
1 / -0

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? The surface tension of mercury at that temperature (20⁰ C) is 4.65 × 10^–1 N m^–1. The atmospheric pressure is 1.01 × 10⁵ Pa.

A
2.30×10⁵ Pa

B
2.01 × 10⁵Pa

C
3.01 × 10⁵ Pa

D
1.01×10⁵ Pa

Solution

Hint: The pressure difference between the concave side and convex side of the spherical surface is called excess pressure.

Step: Calculate pressure inside the drop.

Given,

The radius of mercury drop,r₁= 3.00 mm = 3 × 10⁻³ m

Surface tension, S = 4.65 × 10⁻¹ N m⁻¹

Atmospheric pressure = P₀= 1.01 × 10⁵ cm

Now,

The pressure inside the drop

= Excess pressure + Atmospheric pressure