Splitting & Merging of Drops

The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is [ $T$ = surface tension of the liquid]

A liquid drop having surface energy E is sprayed into $512$ droplets of the same size. Then the final surface energy is

$729$ water droplets of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is ______ - ( $S$ is the surface tension of mercury).

A spherical drop of liquid splits into $1000$ identical spherical drops. If $u_i$ is the surface energy of the original drop and $u_f$ is the total surface energy of the resulting drops, the (ignoring evaporation). $u_f/u_i = (10/x)$. Then value of $x$ is

The work done in splitting a water drop of radius 'R' into 64 droplets is (T is the surface tension of water)

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _______. (Take surface tension of liquid = 0.08 N/m)

Two water drops each of radius $r$ coalesce to form a big drop. If $T$ is the surface tension, the surface energy released in this process is $4\pi r^2 T x$. The value of $x$ is

When a big drop of water is formed from n small drops of water, the energy loss is 3E, where E is the energy of the big drop. If R is the radius of big drop and r is the radius of the smaller drop then number of smaller drops n is

Determine the ratio of surface energy of $1$ large drop to $1$ small drop, if $1000$ small drops combined to form $1$ large drop.

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

A liquid drop having surface energy $E$ is sprayed into 512 droplets of same size. The final surface energy is

The surface energy of a liquid drop is ' $V$ '. It is sprayed into 1000 equal droplets. The surface energy of all the droplets is

A liquid drop having surface energy $E$ is spread into 729 droplets of same size. The final surface energy of the droplets is

A spherical liquid drop splits in to 729 identical spherical drops. If $E$ is the surface energy of the original drop and $U$ is the total surface energy of resulting drops, then $\frac{E}{U} = \frac{1}{x}$. The value of $x$ is

The work done in splitting a water drop of radius R into 64 droplets is ( $T =$ surface tension of water)

The energy needed for breaking a liquid drop of radius ' $R$ ' into 216 droplets, each of radius ' $r$ ' is ' $x$ ' times $TR^2$. The value of ' $x$ ' is [ $T =$ surface tension of the liquid].

The energy needed for breaking a liquid drop of radius ' $R$ ' into ' $n$ ' droplets each of radius ' $r$ ' is $[T =$ surface tension of the liquid]

The amount of work done to break a big water drop of radius ' $R$ ' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be

When a big drop of water is formed from ' $n$ ' small drops of water, the energy loss is ' 3 E ' where ' $E$ ' is the energy of the bigger drop. The radius of the bigger drop is ' R ' and that of smaller drop is ' $r$ ' then the value of ' $n$ ' is

Two water drops each of radius ' $r$ ' coalesce to form a bigger drop. If ' $T$ ' is the surface tension, the surface energy released in this process is:

A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become:

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of big drop is $\frac{10}{x}$. The value of $x$ is ________.

A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ be the total surface energy of 1000 small drops of water and $E_2$ be the surface energy of single big drop of water, then $E_1 : E_2$ is $x : 1$ where $x =$ ________.

When a mercury drop of radius ' $R$ ' splits up into 1000 droplets of radius ' $r$ ', the change in surface energy is ( $T =$ surface tension of mercury)

The work done in splitting a water drop of radius R into 64 droplets is ( $T =$ Surface tension of water)

A small liquid drop of radius $R$ is divided into 27 identical liquid drops. If the surface tension is $T$, then the work done in the process will be:

A big water drop is formed by the combination of ' $n$ ' small water droplets of equal radii. The ratio of the surface energy of ' $n$ ' droplets to the surface energy of the big drop is

Let ' $n$ ' is the number of liquid drops, each with surface energy ' $E$ '. These drops join to form single drop. In this process

1000 small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

Eight small drops of mercury each of radius '$r$', coalesce to form a large single drop. The ratio of total surface energy before and after the change is

A spherical drop of liquid splits into 1000 identical spherical drops. If '$E_1$' is the surface energy of the original drop and '$E_2$' is the total surface energy of the resulting drops, then $\frac{E_1}{E_2} = \frac{x}{10}$. Then value of '$x$' is

A liquid drop of radius '$R$' is broken into '$n$' identical small droplets. The work done is [T = surface tension of the liquid]

A spherical liquid drop of radius $R$ is divided into 8 equal droplets. If surface tension is $S$, then the work done in this process will be

A mercury drop of radius $10^{-3}\text{ m}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45\text{ N m}^{-1}$. The gain in surface energy is:

A spherical drop of liquid splits into 1000 identical spherical drops. If u${}_{i}$ is the surface energy of the original drop and u${}_{f}$ is the total surface energy of the resulting drops, then (ignoring evaporation), $\frac{u_f}{u_i} = \left(\frac{10}{x}\right)$. Then value of x is ____________:

Twenty seven droplets of water each of radius $0.1\text{ mm}$ merge to form a single drop then the energy released is

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

$\left(\text{Take } \pi = \frac{22}{7}\right)$

A large number of water droplets each of radius '$r$' combine to form a large drop of Radius '$R$'. If the surface tension of water is '$T$' & mechanical equivalent of heat is '$J$' then the rise in temperature due to this is

'$n$' number of liquid drops each of radius '$r$' coalesce to form a single drop of radius '$R$'. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is

$[T =$ surface tension of liquid, $\rho =$ density of liquid.]

A water drop of radius $1\text{ cm}$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $\text{cm}$, then the gain in surface energy upto first decimal place will be:

(Given $\pi = 3.14$)

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be:

A water drop of radius $1\text{ cm}$ is broken into eight equal droplets. Surface tension of water is $0.075\text{ N m}^{-1}$. The gain in surface energy is ___________ $\times 10^{-7}\text{ J}$. (Take $\pi = 3.14$)

The surface energy of a liquid drop is 'U'. It splits up into 512 equal droplets. The surface energy becomes

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

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

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

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

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

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

A large number of liquid drops each of radius r coalesce to form a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of liquid T, density $\rho$)

Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T,$ density of liquid is $\rho$ and $L$ is its latent heat of vaporization.