The ratio of the surface tension S and density ρ of liquid 1 and 2 are 1:2 and 1:4 respectively. Equal amount of the two liquids is poured into two identical tubes. what will be the ratio of the rise in the liquid level in the two tubes?

Question

Answer

Problem Statement

Given: Ratio of surface tension S and density ρ of liquid 1 and 2 are 1:2 and 1:4 respectively. Equal amount of the two liquids is poured into two identical tubes. We need to find the ratio of the rise in the liquid level in the two tubes.

Solution

Let us assume that the rise in liquid level in two tubes containing liquid 1 and liquid 2 be h1 and h2 respectively.

Given, ratio of surface tension S and density ρ of liquid 1 and 2 are 1:2 and 1:4 respectively.

Let S1 and S2 be the surface tension of liquid 1 and liquid 2 respectively. Let ρ1 and ρ2 be the densities of liquid 1 and liquid 2 respectively.

It is known that, the pressure inside the liquid at a point below the surface is given by P = P0 + ρgh, where P0 is the atmospheric pressure, h is the depth of the point below the surface and g is acceleration due to gravity.

Now, the rise in liquid level in two tubes containing liquid 1 and liquid 2 be h1 and h2 respectively.

From the above equation, we can write, P1 = P2 (since the two tubes are identical and open to atmosphere at the top)

Therefore, P0 + ρ1gh1 = P0 + ρ2gh2

Canceling out P0 from both sides and rearranging, we get h1/h2 = (ρ2/ρ1) * (S1/S2)

Substituting the given values, we get h1/h2 = (1/2) * (1/4) = 1/8

Therefore, the ratio of the rise in the liquid level in the two tubes is 1:8.

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