Question

A large number of liquid drops each of radius 'r' coalesce to form a big drop of radius 'R'. The energy released in the process in converted into kinetic energy of the big drop. The speed of the big drop is ______. (T = surface tension of liquid, p = density of liquid)

Options

• `["3T"/rho (1/"r" - 1/"R")]^(1/2)`

• `["6T"/rho (1/"r" + 1/"R")]^(1/2)`

• `["6T"/rho (1/"r" - 1/"R")]^(1/2)`

• `["3T"/rho (1/"r" + 1/"R")]^(1/2)`

Answer 1

A large number of liquid drops each of radius 'r' coalesce to form a big drop of radius 'R'. The energy released in the process in converted into kinetic energy of the big drop. The speed of the big drop is `underline(["6T"/rho (1/"r" - 1/"R")]^(1/2))`.

Explanation:

`rho("n" xx (4pi)/3 "r"^3) = rho ((4pi)/3 "R"^3)`

`therefore "n" = "R"^3/"r"^3`

`Delta "A" = 4pi("nr"^2 - "R"^2)`

`= 4pi("R"^3/"r"^3 * "r"^2 - "R"^2)`

`= 4pi"R"^3[1/"r" - 1/"R"]`

∴ E = T Δ A = `4pi "TR"^3 [1/"r" - 1/"R"]`

`1/2 rho (4pi)/3 "R"^3 "v"^2 = 4pi "TR"^3 [1/"r" - 1/"R"]`

`"v"^2 = "6T"/rho(1/"r" - 1/"R")`

v = `["6T"/rho (1/"r" - 1/"R")]^(1/2)`