The JEE (Joint Entrance Examination) is a highly competitive exam in India, known for its rigorous preparation and intense pressure. One of the most challenging aspects of JEE is the Physics section, particularly the topic of fluid mechanics. Among the various problems, the 'dropper who got air' or 'dropping a dropper into water' question has been a recurring theme, often leaving students bewildered.
This seemingly simple problem involves a dropper filled with water being immersed in a beaker of water. When the dropper is released, it sinks to the bottom, and air gets trapped inside. The question typically asks why the dropper floats back up instead of remaining submerged. Let's delve into the physics behind this intriguing phenomenon.
Buoyancy and Archimedes' Principle
To understand why the dropper floats back up, we must first grasp the concept of buoyancy and Archimedes' Principle. Buoyancy is the upward force exerted by a fluid on an object, and Archimedes' Principle states that the upward force exerted by a fluid on an object is equal to the weight of the fluid displaced by the object.
In the case of the dropper, when it is released, it displaces a volume of water equal to its own volume. According to Archimedes' Principle, the dropper experiences an upward force equal to the weight of this displaced water. This force is what we perceive as buoyancy.
Why the Dropper Sinks Initially
Initially, the dropper sinks because the weight of the water inside it is greater than the weight of the water it displaces. This is due to the air trapped inside the dropper, which reduces the overall volume of water displaced. As a result, the buoyancy force is not strong enough to counterbalance the weight of the dropper and its contents.
However, as the dropper sinks, the pressure of the surrounding water increases. This increased pressure forces some of the air out of the dropper, reducing the volume of the trapped air. Consequently, the volume of water displaced by the dropper increases, leading to a stronger buoyancy force.
The Dropper's Ascent
Eventually, the buoyancy force becomes strong enough to counterbalance the weight of the dropper and its contents. At this point, the dropper stops sinking and begins to rise. As it ascends, the pressure of the surrounding water decreases, causing more air to escape from the dropper. This further increases the volume of water displaced, making the buoyancy force even stronger.
The process continues until all the air has escaped from the dropper. Once this happens, the dropper displaces a volume of water equal to its own volume, and the buoyancy force is equal to its weight. At this point, the dropper floats back up to the surface, and the water inside it is in equilibrium with the water outside.
The Role of Pressure and Density
The 'dropper who got air' problem also highlights the relationship between pressure, density, and buoyancy. In a fluid at rest, the pressure increases with depth due to the weight of the overlying fluid. This increase in pressure is what forces the air out of the dropper as it sinks.
Moreover, the density of the fluid also plays a crucial role in buoyancy. In the case of the dropper, the density of the water inside it is initially greater than the density of the surrounding water due to the trapped air. This is why the dropper sinks initially. However, as the air escapes, the density of the water inside the dropper increases, and it begins to rise.
Understanding the physics behind the 'dropper who got air' problem requires a solid grasp of fluid mechanics, buoyancy, and Archimedes' Principle. It's a fascinating example of how seemingly simple phenomena can be explained by fundamental principles of physics. So, the next time you encounter this problem, remember the role of buoyancy and pressure in making the dropper float back up.