Question 129: Properties of Matter

Question

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 :

Accepted Answer

R is the radius of bigger drop.
r is the radius of n water drops.
Water drops are combined to make bigger drop.
So,
Volume of n drops = volume of bigger drop
$$n\left( {{4 \over 3}\pi {r^3}} ight) = {4 \over 3}\pi {R^3}$$
$$ \Rightarrow $$ $$R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} ight)^3}$$
Loss in surface energy, $$\Delta$$U = T $$ 	imes $$ (Change in surface area)
$$\Delta$$U = T (n4$$\pi$$r
2
$$-$$ 4$$\pi$$R
2
)
$$\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} ight)}^3}{r^2} - {R^2}} ight] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} ight)} \over J}$$
$$ 	herefore $$ $${{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} ight)} \over {J 	imes {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} ight]$$

Answer

$${{2T} \over J}\left( {{1 \over r} - {1 \over R}} \right)$$

Answer

$${{3T} \over J}\left( {{1 \over r} - {1 \over R}} \right)$$

Answer

$${{3T} \over rJ}$$

Answer

$${{2T} \over rJ}$$

Properties of Matter

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 :

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