Question

While the piston is at a distance \(2 L\) from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is (A) \(\left(\frac{2 P_{0} \pi R^{2}}{\pi R^{2} P_{0}+M g}\right)(2 L)\) (B) \(\left(\frac{P_{0} \pi R^{2}-M g}{\pi R^{2} P_{0}}\right)(2 L)\) (C) \(\left(\frac{P_{0} \pi R^{2}+M g}{\pi R^{2} P_{0}}\right)(2 L)\) (D) \(\left(\frac{P_{0} \pi R^{2}}{\pi R^{2} P_{0}-M g}\right)(2 L)\)

Short answer

(A) \(\left(\frac{2 P_{0} \times \pi R^{2}}{\frac{R^2}{\pi} \times P_{0}+Mg}\right)(2L)\) is the correct equilibrium distance formula.

Step-by-step solution

Understand the Physics

The problem presents a piston in a cylinder with a hole at the top that is sealed. Initially, the piston is at a distance of 2L from the top, not in equilibrium due to the forces acting on it. When it is released, the piston moves to a new position of equilibrium. At this point, the forces due to air pressure on the top surface of the piston and the gravitational force on the piston must balance out. The air pressure acting upwards is given by the product of the pressure and the area of the piston \(P_0 \times \frac{R^2}{\pi}\), while the weight of the piston acting downwards is given by \(Mg\).

Set Up the Equilibrium Condition

To find the new equilibrium position, we set the downward gravitational force equal to the upward force exerted by the pressure acting on the area of the piston. The equilibrium condition is: \(P_0 \times \pi R^2 = Mg\). This equation states that the forces in opposite directions should be equal for equilibrium to occur.

Solve for the Distance of the Piston from the Top

The distance at which the piston is in equilibrium is not directly affected by the forces at play, but the forces guide us to the correct formula for the distance. Since the problem is asking for the formula rather than a specific value, and we only have the force balance equation, the new distance from the top cannot be directly calculated from it. Rather, we must identify which of the given options corresponds to an equilibrium scenario where the forces balance each other out.

Evaluate the Options for the Correct Formula

We evaluate the given options to find a condition where the forces are balanced. The correct option should simplify to the force balance equation \(P_0 \times \pi R^2 = Mg\).The correct formula, upon simplification, should not have any dimension of length, which is represented by \(L\) because the distance of equilibrium in terms of \(L\) depends on the force balance and not \(L\) itself.

Choose the Correct Option

Option (A) represents the equilibrium distance as \(\left(\frac{2 P_{0} \times \pi R^{2}}{\pi R^{2} \times P_{0}+Mg}\right)(2L)\). If we simplify the fraction inside the parentheses, we see that this matches the equilibrium condition \(P_0 \times \pi R^2 = Mg\), thus it is the correct formula for the distance of the piston from the top when in equilibrium.

Key concepts

Equilibrium Conditions

In physics, particularly when studying mechanics, the concept of equilibrium conditions is essential. Equilibrium is the state of a system where all forces are balanced, and there is no net force causing motion. For an object to be in equilibrium, two main conditions must be met: translational equilibrium, which ensures there is no net force causing linear motion, and rotational equilibrium, which ensures there is no net torque causing rotational motion.

In the case of the piston problem from the JEE Advanced Physics question, we're looking at translational equilibrium. Here, the upward force due to air pressure, given by the product of pressure (\(P_0\)) and the piston's top surface area (\( \frac{R^2}{\theta} \)), needs to be equal to the downward gravitational force, represented by the piston's weight (\(Mg\)). When these two forces equate, the piston will stop moving and find its balance, reaching a state of equilibrium. For students to grasp this, visualizing the scenario can be very helpful. Additionally, thinking of a seesaw in perfect horizontal position due to equally distributed weights can provide a real-world analogy for equilibrium.

Pressure and Force

The relationship between pressure and force plays a critical role in many physics problems. Pressure (\(P\)) is defined as the amount of force exerted per unit area (\(A\) is for area). The formula for pressure is
\[ P = \frac{F}{A} \]
Where
\[ F \text{ is the force } \ A \text{ is the area over which the force is distributed} \]
In this particular JEE Advanced Physics problem involving a piston, pressure is exerted on the piston due to the trapped air. It is crucial for students to understand that this is how the external force of the air causes the piston to move (or stay in place when in equilibrium). The overall force can be calculated by multiplying this pressure by the area (\(P_0 \times \pi R^2\)) the pressure is acting upon. If students are ever confused, reminding them that pressure is like the intensity of force over a certain space—like how high heels can hurt more than sneakers due to the smaller area—can help make the concept more relatable.

Gravitational Force

The gravitational force is a natural phenomenon by which all things with mass or energy are brought towards one another. On Earth, this force is what gives us weight and it acts on all objects at all times. The formula to calculate the gravitational force that an object experiences due to Earth's gravity is given by
\[ F_g = Mg \]
where
\[ M \text{ represents the mass of the object } \ g \text{ is the acceleration due to gravity (approximately } 9.8 \text{ m/s}^2 \text{ on Earth) } \]
For students solving the piston problem, understanding the impact of gravitational force is vital in predicting the piston’s behavior. The gravitational force on the piston (\(Mg\)) causes it to move downwards, seeking the point of equilibrium against the upward pressure force. It’s similar to the downward pull we feel that keeps us grounded on Earth. A helpful tip for students might be to imagine holding a ball in the air and then letting it go—the ball falls due to the gravitational force, just as the piston will move until equilibrium is achieved.