Question

Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is ____________.

Options

• `sqrt(2) : 1`

• `2^(1/3) : 1`

• 2 : 1

• `2^(2/3) : 1`

Answer 1

Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is `underlinebb(2^(1/3) : 1)`.

Explanation:

When two small drops of mercury coalesce to form a large drop, then volume remains constant.

`2 xx (4π"r"^3)/3 = 4/3 π"R"^3`

R = `2^(1/3)"r"` ........(i)

The total surface energy before change is

U₁ = 2 × T × ΔA = 2 × T × 4πr² .........(ii)

and total surface energy after change is

U₂ = T × ΔA = T × 4πR² .........(iii)

Dividing Eqs. (ii) by (iii), we get

`("U"_1)/("U"_2) = (2 xx "T" xx 4π"r"^2)/("T" xx 4π"R"^2) = (2"r"^2)/(2^(2/3)"r"^2)` ......[using Eq. (i)]

= `(2^(1/3))/1` or `2^(1/3) : 1`