We know that surface acting agents (i.e. surfactants) such as soaps and detergents belong to the class of micelles. A miceller system when dissolved in water, dissociates to give ions. The anion consists of two parts. The polar groups such as $$\left( {CO{O^ - }\,{\text{or}}\,\,SO_4^{ - - }} \right)$$    ion is water loving (i.e. hydrophilic) in nature. It is called head of the species. The hydrocarbon chain which is quite big in size is water repelling (i.e. hydrophobic) in nature. It is called tail of the species. The hydrocarbon chain aggregates into the micelle above the critical concentration.
NOTE : It may also be noted that the critical concentration for micelle formation decreases with increase in the molecular weight of the hydrocarbon chain of surfactant.
Here the two anions that are formed are in case of "B"
$$\left( {{\text{i}}{\text{.e}}{\text{.}}\,\,C{H_3}{{\left( {C{H_2}} \right)}_{11}}OSO_3^ - } \right)\,$$     and "C" $$\left( {{\text{i}}{\text{.e}}.\,\,C{H_3}{{\left( {C{H_2}} \right)}_6}{\text{CO}}\mathop {\text{O}}\limits^ - } \right)$$
The molecular weight of hydrocarbon chain is more in case of "B" so it has lower value of critical concentration for micelle formation in aqueous solution.
Hence the correct answer is option (b).

In multimolecular colloids, the smaller particles aggregate and are held together by van der Waals' forces, e.g. sols of gold atoms and sulphur molecules.

Tyndall effect is optical property, whereas other properties are electrical properties. Hence dependent on the charge on colloids

$$\eqalign{ & {\text{Volume of gold dispersed in }}1{\text{ }}L{\text{ water}} \cr & {\text{ = }}\frac{{mass}}{{Density}} = \frac{{1.9 \times {{10}^{ - 4}}g}}{{19gm\,c{m^{ - 3}}}} = 1 \times {10^{ - 5}}\,c{m^3} \cr & {\text{Radius of gold sol particle}} \cr & = 10\,nm = 10 \times {10^{ - 7}}\,cm = {10^{ - 6}}\,cm \cr & {\text{Volume of gol sol particle}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3} \times \frac{{22}}{7} \times {\left( {{{10}^{ - 6}}} \right)^3} = 4.19 \times {10^{ - 18}}c{m^3} \cr & \therefore \,\,{\text{No}}{\text{. of gold sol particles in}}\,\,1 \times {10^{ - 5}}c{m^3} \cr & = \frac{{1 \times {{10}^{ - 5}}}}{{4.19 \times {{10}^{ - 18}}}} = 2.38 \times {10^{12}} \cr & \therefore \,\,{\text{No}}{\text{. of gold sol particles in one}}\,m{m^3} \cr & = \frac{{2.38 \times {{10}^{12}}}}{{{{10}^6}}} \cr & = 2.38 \times {10^6} \cr} $$

More easily liquefiable gases are adsorbed readily. Thus, $${H_2}$$  gas having low critical temperature $$\left( {33\,K} \right)$$  is not easily liquefied and shows least adsorption.

For a protective colloid $$\mu $$ lesser the value of gold number better is the protective power.
Thus the correct order of protective power of A, B, C and D is
$$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( A \right)\,\,\, < \,\,\left( C \right)\,\,\,\, < \,\,\left( B \right)\,\, < \,\,\left( D \right) \cr & {\text{Gold number}}\,\,\,\,{\text{0}}{\text{.50}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}}{\text{.10}}\,\,\,\,\,\,\,\,\,\,\,{\text{0}}{\text{.01}}\,\,\,\,\,\,\,\,\,\,\,{\text{0}}{\text{.005}} \cr} $$
Hence (C) is the correct answer

Rate of physiorsption increases with decrease of temperature

$$\frac{x}{m} = $$   amt of gas adsorbed per unit mass of adsorbent
$$t=$$  temperature

Tyndall effect is due to the scattering of light by colloidal particles. Sugar solution is a true homogeneous solution which will not show Tyndall effect.

In gel, dispersed phase is liquid and disperse medium is solid.

At $$CMC,$$  the particles of an electrolyte aggregate and form associated colloids known as micelles.