Q. No.If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms a cylindrical surface on the side that gives rise to lower pressure in the water in comparison to the atmosphere. If the radius of the cylindrical surface is \(R\) and the surface tension of water is \(T\) then the pressure in water between the plates is lower by:
1. \(\frac{2{T}}{{R}}\)
2. \(\frac{4{T}}{{R}}\)
3. \(\frac{{T}}{4{R}}\)
4. \(\frac{{T}}{{R}}\)
1. \(\frac{2{T}}{{R}}\)
2. \(\frac{4{T}}{{R}}\)
3. \(\frac{{T}}{4{R}}\)
4. \(\frac{{T}}{{R}}\)
Subtopic: Surface Tension |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
The pressure inside two soap bubbles are \(1.01\) and \(1.02\) atmosphere, respectively. The ratio of their volumes is:
1. \(4:1\)
2. \(2:1\)
3. \(8:1\)
4. \(0.8:1\)
Subtopic: Surface Tension |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A large number of water drops, each of radius \(r,\) combine to have a drop of radius \(R.\) If the surface tension is \(T\) and mechanical equivalent of heat is \(J\), the rise in heat energy per unit volume will be:
| 1. | \(\dfrac{2T}{J}\left (\dfrac{1}{r}- \dfrac{1}{R}\right)\) | 2. | \(\dfrac{2T}{Jr}\) |
| 3. | \(\dfrac{3T}{Jr}\) | 4. | \(\dfrac{3T}{J}\left (\dfrac{1}{r}- \dfrac{1}{R}\right)\) |
Subtopic: Surface Tension |
66%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be:
1. \(2.8 \times 10^{-4} \text J\)
2. \(1.5 \times 10^{-3} \text J\)
3. \(1.9 \times 10^{-4} \text J\)
4. \(9.4 \times 10^{-5} \text J\)
1. \(2.8 \times 10^{-4} \text J\)
2. \(1.5 \times 10^{-3} \text J\)
3. \(1.9 \times 10^{-4} \text J\)
4. \(9.4 \times 10^{-5} \text J\)
Subtopic: Surface Tension |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A water drop of a radius \(1\) cm is broken into \(729\) equal droplets. If the surface tension of water is \(75\) dyne/cm, then the gain in surface energy upto first decimal place will be:
[Given \(\pi =3.14\)]
1. \(8.5 \times 10^{-4} \) J
2. \(8.2 \times 10^{-4} \) J
3. \(7.5 \times 10^{-4} \) J
4. \(5.3 \times 10^{-4} \) J
[Given \(\pi =3.14\)]
1. \(8.5 \times 10^{-4} \) J
2. \(8.2 \times 10^{-4} \) J
3. \(7.5 \times 10^{-4} \) J
4. \(5.3 \times 10^{-4} \) J
Subtopic: Surface Tension |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
Given below are two statements:
| Assertion (A): | Clothes containing oil or grease stains cannot be cleaned by water wash |
| Reason (R): | Because the angle of contact between the oil/grease and water is obtuse. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
Subtopic: Surface Tension |
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A big drop is divided into \(1000\) identical droplets. If the big drop had surface energy \(U_{i}\) and all small droplets together had a surface energy \(U_{f},\) then \(\dfrac {U_i}{U_f}\) is equal to:
1. \(1/100\)
2. \(10\)
3. \(1/10\)
4. \(1000\)
1. \(1/100\)
2. \(10\)
3. \(1/10\)
4. \(1000\)
Subtopic: Surface Tension |
63%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
The excess pressure inside a soap bubble of radius \(R\) and surface tension \(T\) is:
| 1. | \(\dfrac{T}{R}\) | 2. | \(\dfrac{2T}{R}\) |
| 3. | \(\dfrac{3T}{R}\) | 4. | \(\dfrac{4T}{R}\) |
Subtopic: Surface Tension |
80%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
The amount of work required to expand the soap bubble from a radius of \(r_1=3.5~\text{cm}\) to \(r_2=7.0~\text{cm}\) is:
(given surface tension of soap solution, \(T=0.03~\text{N/m}\))
1. \(0.14~\text{mJ}\)
2. \(1.4~\text{mJ}\)
3. \(0.7~\text{mJ}\)
4. \(2.8~\text{mJ}\)
Subtopic: Surface Tension |
55%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
A drop of water of \(10\) mm radius is divided into \(1000\) droplets. If the surface tension of the water surface is equal to \(0.073 \) J/m2 then increment in surface energy while breaking down the bigger drop in small droplets as mentioned is equal to:
1. \(8.25 \times 10^{-5} \mathrm{~J}\)
2. \(9.17 \times 10^{-4} \mathrm{~J}\)
3. \(9.17 \times 10^{-5} \mathrm{~J}\)
4. \(8.25 \times 10^{-4} \mathrm{~J}\)
1. \(8.25 \times 10^{-5} \mathrm{~J}\)
2. \(9.17 \times 10^{-4} \mathrm{~J}\)
3. \(9.17 \times 10^{-5} \mathrm{~J}\)
4. \(8.25 \times 10^{-4} \mathrm{~J}\)
Subtopic: Surface Tension |
57%
From NCERT
JEE
Please attempt this question first.
Hints
Please attempt this question first.
© 2025 GoodEd Technologies Pvt. Ltd.