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Surface Energy - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Surface energy is considered one the most difficult concept.

  • 15 Questions around this concept.

Solve by difficulty

The surface tension of a soap bubble is 2.0 × 10–2 Nm–1. Work done to increase the radius of the soap bubble from 3.5cm to 7 cm will be:
Take \left [\pi =\frac{22}{7} \right ]

If 1000 droplets of water of surface tension \mathrm{0.07 N/m}, having the same radius \mathrm{1 mm} each, combine to form a single drop. In the process, the released surface energy is-

\mathrm{(Take\: \pi=\frac{22}{7})}

A mercury drop of radius 10^{-3}m  is broken into 125 equal-sized droplets.
The surface tension of mercury is  0.45Nm^{-1} . The gain in surface energy is :

The work done in blowing a soap bubble of volume 'v' is w. The work done in blowing a soap bubble of volume '2 v' is.

Two small drops of mercury each of radius r from a single large drop. The ratio of surface energy before and after this change is:

The energy of a bubble is E. If one thousand such bubbles coalesce to form a large bubble, its energy is E1. If the ratio of E and E1 will be equal to  \frac{1}{x}, the value of x is.

Concepts Covered - 1

Surface energy

The molecules on the liquid surface experience net downward force. And because of this force, these molecules tend to move downwards. So to fill the space we need to bring a molecule from the interior of the liquid to the free surface. And to do this some work is required to be done against the intermolecular force of attraction. This work will be stored as the potential energy of the molecule on the surface. 

And this stored potential energy of surface molecules per unit area of the surface is called surface energy.

Surface energy is also defined as the amount of work done in increasing the area of the surface film through unity.

$
\begin{aligned}
& \text { I.e surface energy }=\frac{\text { work done in increasing the surface area }}{\text { increase in surface area }} \\
& \text { or surface energy }=\frac{W}{\Delta A} \ldots \text { (1) }
\end{aligned}
$


Where $W \rightarrow$ work done

$
\text { and } \Delta A \rightarrow \text { increase in area }
$


And work done in increasing the surface area is given by

$
W=T \times \Delta A \ldots . .(2)
$

where $T \rightarrow$ Surface tension

$
\text { and } \Delta A \rightarrow \text { increase in area }
$


So we rewrite equation (2) as

$
T=\frac{W}{\Delta A} \ldots(3)
$


So we can also define surface tension as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension.
Or we can say that the surface tension of a liquid is numerically equal to its surface energy.
As

$
W=T \Delta A
$


If $\Delta A=1$, then $T=W$
$T \rightarrow$ Surface tension

 

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Surface energy

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