Tag: surface energy of a liquid

Questions Related to surface energy of a liquid

A spherical water drop of radius $R$ is split up into $8$ equal droplets. If $T$ is the surface tension of water, then the work done in this process is-

  1. $4\pi R^2T$

  2. $8\pi R^2T$

  3. $48\pi R^2T$

  4. $2\pi R^2T$


Correct Option: A
Explanation:

Let smaller drop has radius $r$

Volume will remain same. 
$\dfrac{4}{3}\pi R^3 = 8\times \dfrac{4}{3} \pi r^3  \implies   r= \dfrac{R}{2}$

Change in surface energy-
 $A _2T - A _1T = 8\times 4\pi r^2T-4\pi R^2T=4\pi[2R^2-R^2]T=4\pi R^2T$
Work done  = change in surface energy  = $4\pi R^2T$

A drop of oil is placed on the surface of water. Which of the following statement is correct?

  1. It will remain on it as a sphere

  2. It will spread as a thin layer

  3. It will partly be as spherical droplets and partly as thin film

  4. It will float as distorted drop on the water surface.


Correct Option: B
Explanation:

Adhesive force between oil and water molecules is greater than cohesive force between oil molecules. So the oil molecules do not mix with water molecules. As a result, oil spreads on the water surface forming a thin layer.

One thousand small water droplets of equal size combine to form a big drop. The ratio of the final surface energy to the initial surface energy is: (Surface tension of water = 70 dyne/cm) 

  1. 10:1

  2. 1000:1

  3. 1:10

  4. 1:1000


Correct Option: C
Explanation:

Let the radius of smaller radius be $r$ and bigger be $R$.

Since volume remains constant, thus  $\dfrac{4\pi R^3}{3}=1000\times \dfrac{4\pi r^3}{3}$
We get  $R=10r$
Ratio of final surface energy to initial is 
$\dfrac{W _f}{W _i}=\dfrac{T\times 4\pi R^2}{T\times 1000\times 4\pi r^2}$
$\dfrac{W _f}{W _i}=\dfrac{(10r)^2}{1000r^2}=1:10$

A metal plane having an area of $0.04\ m^{2}$ is placed on a horizontal wooden surface. Oil of coefficient of viscosity $2\ N/ s/m^{2}$ is introduced between the plate and the surface till the thickness of the oil layer is $0.5$ in. The horizontal force needed to drag the plate along the surface with a velocity of $5\ cm/s$ is  

  1. $80\ N$

  2. $8\ N$

  3. $60\ N$

  4. $6\ N$


Correct Option: A

Two spherical soap bubbles of a radii $r _1$ and $r _2$ in vacuum coalesce under isothermal conditions. The resulting bubble has the radius $R$ such that

  1. $R=r _1+r _2$

  2. $R^2 ={ r _1^2 + r _2^2}$

  3. $R=\dfrac{r _1+r _2}{r _2}$

  4. none of these


Correct Option: B

What is the change in surface energy, when a mercury drop of radius $R$ splits up into $1000$ droplets of radius $r$?

  1. $8\pi R^2T$

  2. $16\pi R^2T$

  3. $24\pi R^2T$

  4. $36\pi R^2T$


Correct Option: D
Explanation:

Let Surface Tension be T.

Initial surface = $4\Pi { R }^{ 2 }$
Final Surface area = $1000\times 4\Pi { r }^{ 2 }$
& Initial Volume = final volume
$\dfrac { 4 }{ 3 } \Pi { R }^{ 3 }=1000\times \dfrac { 4 }{ 3 } \Pi { r }^{ 3 }$
R = 10r
So,
    Final surface area = $1000\times 4\Pi \times \dfrac { { R }^{ 2 } }{ 100 } =40\Pi { R }^{ 2 }$
So, Workdone = Change in Surface energy
                         = $T\Delta A=(40-4)\Pi { R }^{ 2 }T$
       $\boxed { Change\quad in\quad Surface\quad energy=36\Pi { r }^{ 2 }T } $.

Potential energy of a molecule on the surface of a liquid is as compared to another molecule inside of the liquid is

  1. More

  2. Less

  3. Both a and b

  4. None of these


Correct Option: A
Explanation:

Due to surface Tension, Potential energy of a molecule at free surface is more than the molecules inside the liquid.