Two Questions from Properties of Matter:
The following question appeared in the Kerala Engineering Entrance test paper of 2006:
The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere respectively. The ratio of their respective volumes is
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
To the surprise of this author, a comparatively bright student omitted this question, which is quite simple. You know that the excess of pressure inside a bubble is 2T/r, where T is the surface tension and r is the radius. The ratio of the excess pressures inside the bubbles is P1/P2 = r2/r1. But, P1/P2 =0.01/0.02. [Since the actual pressures are 1.01 and 1.02 atmosphere]
So, we have r2/r1 = 1/2. The ratio of volumes, V1/V2 = (r1/r2)^3 = 2^3 = 8.
Consider now the following question which appeared in Kerala Medical Entrance test paper of 2006:
To what depth below the surface of sea should a rubber ball be taken so as to decrease its volume by 0.1%? [Take: Density of sea water = 1000kg/m^3, bulk modulus of rubber = 9×10^8 N/m^2, acceleration due to gravity = 10m/s^2]
(a) 9m (b) 18m (c) 180m (d) 90m (e) 900m
We have bulk modulus B = P/(dv/v) so that P = B(dv/v) = (9×10^8) ×0.001 = 9×10^5 pascal.
Therefore, hdg = 9×10^5 from which, h = (9×10^5)/(1000×10) = 90m. [Option (d)]
The following question appeared in the Kerala Engineering Entrance test paper of 2006:
The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere respectively. The ratio of their respective volumes is
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
To the surprise of this author, a comparatively bright student omitted this question, which is quite simple. You know that the excess of pressure inside a bubble is 2T/r, where T is the surface tension and r is the radius. The ratio of the excess pressures inside the bubbles is P1/P2 = r2/r1. But, P1/P2 =0.01/0.02. [Since the actual pressures are 1.01 and 1.02 atmosphere]
So, we have r2/r1 = 1/2. The ratio of volumes, V1/V2 = (r1/r2)^3 = 2^3 = 8.
Consider now the following question which appeared in Kerala Medical Entrance test paper of 2006:
To what depth below the surface of sea should a rubber ball be taken so as to decrease its volume by 0.1%? [Take: Density of sea water = 1000kg/m^3, bulk modulus of rubber = 9×10^8 N/m^2, acceleration due to gravity = 10m/s^2]
(a) 9m (b) 18m (c) 180m (d) 90m (e) 900m
We have bulk modulus B = P/(dv/v) so that P = B(dv/v) = (9×10^8) ×0.001 = 9×10^5 pascal.
Therefore, hdg = 9×10^5 from which, h = (9×10^5)/(1000×10) = 90m. [Option (d)]
