Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $$4: 3$$. The rise of liquid in two capillaries is '$$h_1$$' and '$$h_2$$' respectively. If the surface tensions of liquids are in the ratio $$6: 5$$, the ratio of heights $$\left(\frac{h_1}{h_2}\right)$$ is

(Assume that their angles of contact are same)

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

A body of density '$$\rho$$' is dropped from rest at a height '$$h$$' into a lake of density '$$\sigma' (\sigma>\rho)$$. The maximum depth to which the body sinks before returning to float on the surface is (neglect air dissipative forces)

'$$n$$' 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

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