Mechanical Properties of Fluids Test

Result

Mechanical Properties of Fluids Test - 72

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Question 1
4 / -1

A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal condition. If T is the surface tension of soap solution, the energy spent in the blowing, is

A
3 πTr²

B
6 πTr²

C
12 πTr²

D
24 πTr²

Solution
Question 2
4 / -1

The surface tension of a liquid is 5 N/m . If a thin film of the area 0.02 m² is formed on a loop, then its surface energy will be

A
5 × 10² J

B
2.5 × 10^–2 J

C
2 × 10^–1 J

D
5 × 10^–1 J

Solution
Question 3
4 / -1

The work done in blowing a soap bubble of radius r of the solution of surface tension T will be

A
8πr²T

B
2πr²T

C
4πr²T

D
Question 4
4 / -1

Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies and before and after the change is

A
1 : 2^1/3

B
2^1/3 :1

C
2 : 1

D
1 : 2

Solution
Question 5
4 / -1

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m²)

A
23.4 µ J

B
18.5 µ J

C
26.8 µ J

D
16.8 µ J

Solution
Question 6
4 / -1

The surface tension of a liquid at its boiling point

A
Becomes zero

B
Becomes infinity

C
Is equal to the value at room temperature

D
Is half to the value at the room temperature
Question 7
4 / -1

What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large drop

A
100 : 1

B
1000 : 1

C
10 : 1

D
1 : 100

Solution
Question 8
4 / -1

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

A
1000 : 1

B
1 : 1000

C
10 : 1

D
1 : 10

Solution
Question 9
4 / -1

The work done in increasing the size of a soap film from 10 cm × 6 cm to 10 cm × 11 cm is 3 × 10^–4 joule. The surface tension of the film is

A
1.5 × 10^–2 N/m

B
3.0 × 10^–2 N/m

C
6.0 × 10^–2 N/m

D
11.0 × 10^–2 N/m

Solution
Question 10
4 / -1

If σ be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius is

A
Rn^2/3 σ

B
(n^2/3 – 1)R σ

C
(n^1/3 – 1)R σ

D
4πR²(n^1/3 – 1) σ