1. Work done in splitting a drop of water of 1 mm radius into 10^{6} droplets is (Surface tension of
water = \[72\times10^{-3} J\diagup m^{2}\] )

a) \[9.58\times10^{-5} J\]

b) \[8.95\times10^{-5} J\]

c) \[5.89\times10^{-5} J\]

d) \[5.98\times10^{-6} J\]

Explanation:

2. A spherical liquid drop of radius R is divided into eight equal droplets. If surface tension is T, then the work done in this process will be

a) \[2\pi R^{2} T\]

b) \[3\pi R^{2} T\]

c) \[4\pi R^{2} T\]

d) \[2\pi R T^{2}\]

Explanation:

3. If T is the surface tension of soap solution, the amount of work done in blowing a soap bubble from a diameter D to 2D is

a) \[2\pi D^{2} T\]

b) \[4\pi D^{2} T\]

c) \[6\pi D^{2} T\]

d) \[8\pi D^{2} T\]

Explanation: Work done to increase the diameter of bubble from d to D

4. The radius of a soap bubble is increased from \[\frac{1}{\sqrt{\pi}} cm\] to \[\frac{2}{\sqrt{\pi}} cm\] . If the surface tension of water
is 30 dynes per cm, then the work done will be

a) 180 ergs

b) 360 ergs

c) 720 ergs

d) 960 ergs

Explanation:

5. If work W is done in blowing a bubble of radius R from a soap solution, then the work done in blowing a bubble of radius 2R from the same
solution is

a) W/2

b) 2W

c) 4W

d) \[2\frac{1}{3} W\]

Explanation:

6. A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. If the surface tension of oil is 50 dynes/cm, the work done is

a) \[18\pi\] ergs

b) \[180\pi\] ergs

c) \[1800\pi\] ergs

d) \[8000\pi\] ergs

Explanation:

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

a) \[8\pi r^{2} T\]

b) \[2\pi r^{2} T\]

c) \[4\pi r^{2} T\]

d) \[\frac{4}{3}\pi r^{2} T\]

Explanation: \[8\pi r^{2} T\]

8. If two identical mercury drops are combined to form a single drop, then its temperature will

a) Decrease

b) Increase

c) Remains the same

d) None of the above

Explanation: Surface energy of combined drop will be lowered, so excess surface energy will raise the temperature of the drop

9.If the surface tension of a liquid is T, the gain in surface energy for an increase in liquid surface by A is

a) \[AT^{-1}\]

b) \[AT\]

c) \[A^{2}T\]

d) \[A^{2}T^{2}\]

Explanation: Surface energy = surface tension × increment in area

= T × A

10. The surface tension of a soap solution is \[2\times10^{-2}N\diagup m\] . To blow a bubble of radius 1 cm,
the work done is

a) \[4\pi\times10^{-6}J\]

b) \[8\pi\times10^{-6}J\]

c) \[12\pi\times10^{-6}J\]

d) \[16\pi\times10^{-6}J\]

Explanation: