MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Surface Energy

Eight droplet...

Question

Eight droplets of mercury, each of radius $1 m m$ , coalesce into a single drop.
(a) Find the radius of the single drop?
[Surface tension of mercury $= 0.465 J / m^{2}$ ].

Open in App

Solution

$n =$ number of droplets $= 8$
$r =$ radius of droplets $= 1$ mm
surface tension $T = 0.465 J / m^{2}$
$r = 1$ mm
$= 1 \times 0.001$ m
radius of single drop $= 0.001$ m

Suggest Corrections

thumbs-up

0

Similar questions

Q. A drop of mercury of radius

2 m m

is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is

0.465 {J / m}^{2}

)

Q. Calculate the work done in breaking a mercury drop of radius

1 m m

1000

droplets of the same size. Surface tension of mercury is

525 \times 10^{- 3} N / m

Q. Eight identical spherical mercury drops charged to a potential of

20 V

each are coalesced into a single spherical drop.

Q. Two small drops of mercury, each of radius R coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

Q. Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is :-

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Surface Tension

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon