Question

The excess pressure inside a spherical drop of
water is frour times that of another drop. Then their
respective mass ratio is
(1) 1 : 16
(2) 8:1
(3) 1:4
(4) 1: 64

Answer 1

Answer: option (4): 1 : 64

Explanation:

Step 1:

Let the excess pressure of water drop be “P₁” and another drop be “P₂”. Also, let the radius of the water and another drop be denoted as “r₁” & “r₂” respectively.

It is given that, P₁ = 4 * P₂ …. (i)

The formula for the excess pressure of a spherical drop is given as,

P = [2 γ] / r …. (ii)

Where

γ = surface tension of the drop

R = radius of the drop

Therefore, from (i) & (ii), we get

[2 γ] / r₁ = 4 * [(2 γ) / r₂]

⇒ r₂ = 4 * r₁ …. (iii)

Step 2:

We know,  

Mass(m) = Volume(V) * density(ρ)

Where  

Volume of a spherical drop = 4/3 πr³

Let mass of water drop be “m₁” & mass of another drop be “m₂”.

Therefore, we have

m₁/m₂ = [4/3πr₁³ * ρ] / [4/3πr₂³ * ρ]

cancelling the similar terms

⇒ m₁/m₂ = [r₁³ / r₂³]

⇒ m₁/m₂ = [r₁/(4*r₁)]³ .... [from (iii)]

⇒ m₁/m₂ = [1/4]³

⇒ m₁/m₂ = 1/64

Thus, the respective ratio of their masses 1:64.