JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 4)
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 :
$${{2T} \over J}\left( {{1 \over r} - {1 \over R}} \right)$$
$${{3T} \over J}\left( {{1 \over r} - {1 \over R}} \right)$$
$${{3T} \over rJ}$$
$${{2T} \over rJ}$$
Explanation
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
$$n\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$$
$$ \Rightarrow $$ $$R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} \right)^3}$$
Loss in surface energy, $$\Delta$$U = T $$ \times $$ (Change in surface area)
$$\Delta$$U = T (n4$$\pi$$r2 $$-$$ 4$$\pi$$R2)
$$\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} \right)}^3}{r^2} - {R^2}} \right] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over J}$$
$$ \therefore $$ $${{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over {J \times {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} \right]$$
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
$$n\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$$
$$ \Rightarrow $$ $$R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} \right)^3}$$
Loss in surface energy, $$\Delta$$U = T $$ \times $$ (Change in surface area)
$$\Delta$$U = T (n4$$\pi$$r2 $$-$$ 4$$\pi$$R2)
$$\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} \right)}^3}{r^2} - {R^2}} \right] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over J}$$
$$ \therefore $$ $${{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over {J \times {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} \right]$$
Comments (0)
