Tag: physics

Questions Related to physics

W is the work done, when a bubble of volume V is formed from a solution. How much work is required to be done to form a bubble of volume 2V?

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $2W$

  2. $W$

  3. $2^{1/3}W$

  4. $2^{2/3}W$

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Correct Option: D
Explanation:

Let the radius of the bubble with volume $V$ be $  r$.

Then the radius of the bubble  with volume $2V$ is $2^{1/3}r$.
Work is only because of change in surface energy.
Ratio of work done =  Ratio of surface energy =Rratio of surface area 
$\dfrac{W _1}{W} = \dfrac{4\pi (2^{1/3}r)^2}{4\pi r^2} \implies W _1= 2^{2/3}W$

The surface energy of a liquid drop of radius r is proportional to:

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $r^3$

  2. $r^2$

  3. r

  4. $\dfrac {1}{r}$

Show answer
Correct Option: B
Explanation:

 as we know that surface energy is proportional to surface area.

${E}\propto{A}\propto{r}^{2}$

A mercury drop of radius 1 cm is broken into 106 droplets of equal size. The work done is

$(T = 35 10^{-2} N/m)$

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $1.615\times 10^{-3}J$

  2. $4.35\times 10^{-4}J$

  3. $1.615\times 10^{-6}J$

  4. $4.35\times 10^{-8}J$

Show answer
Correct Option: A
Explanation:

Let the radius of smaller drop be r, so

$\dfrac{4\pi (1)^2}{3}=106\dfrac{4\pi r^3}{3}$
$r=0.21cm=0.21\times 10^{-2}m$
work done is $W=T(dA)=T(106\times 4\pi (0.21\times 10^{-2})^2-4\pi (1\times 10^{-2})^2)$
here $T=35\times 10^{-2}$
$W=1615\times 10^{-6}J$

Surface tension of a soap solution is $1.9 10^{-2} N/m.$ Work done in blowing a bubble of 2.0cm diameter will be

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $7.6 10^{-6} J$

  2. $15.2 10^{-6} J$

  3. $16 10^{-6} J$

  4. $2.5 10^{-6} J$

Show answer
Correct Option: B
Explanation:

Work done = Change in surface energy

                   = Surface Tension $\times $ Area
                   $= 1.910 \times $ $4\pi \times { \left( 2\times { 10 }^{ -2 } \right)  }^{ 2 }$
workdone = $15.2\times { 10 }^{ -6 } J$.

The surface tension of a soap solution is $0.035 N/m$. the energy needed to increase the radius of the bubble from $4$ cm to $6$ cm is

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $1.75\times 10^{-3} J$

  2. $1.510^{-2} J$

  3. $310^{-3} J$

  4. $1.510^{-4} J$

Show answer
Correct Option: A
Explanation:

There are two free surface available in bubble.

Change in surface area $\Delta A = 2(4\pi R _f^2 - 4\pi R _i^2)$
Energy needed to change radius from 4 to 6 is 

$W=T \Delta A=T\times 2(4\pi R _f^2-4\pi R _i^2)$
$W=.035\times 2\times 4\pi((6\times 10^{-2})^2-(4\times 10^{-2})^2)$
$W=1.75\times 10^{-3}J$

Two drops of a liquid are merged to from a single drop. In this process-

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. Energy is released

  2. Energy is absorbed

  3. Energy remains constant

  4. First B then A

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Correct Option: A
Explanation:

It is nature of free surface to always minimize the surface tension; Hence when two bubble fuse then energy is released due to decrease in surface Tension.

Two soap bubbles of radii 4 cm and 3 cm respectively coalesce under isothermal conditions to form a single bubble. What is the radius of the new single bubble?

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $3 cm$

  2. $4 cm$

  3. $5 cm$

  4. $6 cm$

Show answer
Correct Option: C
Explanation:
Given:
Radius of first soap bubble, $r _1 = 4 cm$
Radius of second soap bubble, $r _2 = 3 cm$
Let,
$P _1 = \dfrac{4T}{(r _1)^2}, V _1 = \dfrac{4}{3} \pi (r _1)^3$ be the excess pressure inside first soap bubble and volume of first soap bubble respectively.  
$P _2 = \dfrac{4T}{(r _2)^2}, V _2 = \dfrac{4}{3} \pi (r _2)^3$ be the excess pressure inside second soap bubble and volume of second soap bubble respectively. 
And $P = \dfrac{4T}{r^2}, V _2 = \dfrac{4}{3} \pi r^3$be the excess pressure inside new soap bubble, volume and radius of new soap bubble respectively. 
The two bubbles combine isothermally hence,
$PV = P _1 V _1 + P _2 V _2$

$\dfrac{4T}{r} \dfrac{4}{3} \pi r^3 = \dfrac{4T}{(r _1)} \dfrac{4}{3} \pi (r _1)^3 + \dfrac{4T}{(r _2)} \dfrac{4}{3} \pi (r _2)^3$

$r^2 = (r _1)^2 + (r _2)^2$
$r = \sqrt ((r _1)^2 + (r _2)^2)$
$r = \sqrt ((4)^2 +(3)^2)$
$r = \sqrt 25$
$r = 5 cm$

A square frame of length $L$ is immersed in soap solution and taken out. The force experienced by the square plate is

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $TL$

  2. $2TL$

  3. $4TL$

  4. $8TL$

Show answer
Correct Option: D
Explanation:

As force applied by tension will be from outer side and from inner side as well.

Total length $L' = 2(4L) = 8L$
So  $F = L'T =  8TL$

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-

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  1. $4\pi R^2T$

  2. $8\pi R^2T$

  3. $48\pi R^2T$

  4. $2\pi R^2T$

Show answer
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?

surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

  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.

Show answer
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.