Final answer:
Assuming similar conditions for surface tension and contact angle, a liquid with twice the density of water will rise to half the height of what water rises in a capillary tube. Therefore, for a liquid with a density of 2 g/cm³, the rise would be 2 cm, given that water rises 4 cm under the same conditions.
Step-by-step explanation:
The phenomenon of capillary action is observed when a liquid is able to rise or fall within a narrow tube, defying the effects of gravity due to interaction between the liquid's molecules and the tube's surface. In this scenario, the rise of a different liquid in a capillary tube with a different density than water is being examined. Factors such as the surface tension of the liquid, the angle of contact between the liquid and the surface, the radius of the capillary tube, and the density of the liquid come into play to affect the capillary rise.
According to the Jurin's law:
H = (2 * S * cos(θ)) / (ρ * g * r)
where H is the height the liquid rises, S is the surface tension, θ is the angle of contact, ρ is the density of the liquid, g is the acceleration due to gravity, and r is the radius of the tube. When considering a liquid other than water with a density of 2 g/cm³ without any other given variables such as surface tension or angle of contact, and assuming these are similar to water, we can simplify the calculation to a function of density because the rise is inversely proportional to density. As the other liquid has double the density of water, it would rise to half the height of water under the same conditions, provided the surface tension and angle of contact are the same. Therefore, if water rises 4 cm, the other liquid would rise to 2 cm.