The pans of a physical balance are in equilibrium. Air is blown under the right hand pan; then the right hand pan wil

Correct Option: Move down

To understand why the right-hand pan moves down, we can apply Bernoulli's principle from fluid dynamics.

Bernoulli's principle states that for an incompressible, non-viscous fluid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in static pressure. Mathematically, along a horizontal streamline, the relationship is given by:
$P+\frac{1}{2}\rho v^2=\text{constant}$
where:
• $P$ is the static pressure of the fluid,
• $\rho$ is the density of the fluid, and
• $v$ is the velocity of the fluid flow.

Let's analyze the forces acting on the right-hand pan step-by-step:
1. Initial State: The pans of the physical balance are in equilibrium, meaning the forces acting on both sides are balanced, and the pressure above and below both pans is equal to the atmospheric pressure ($P_{\text{atm}}$).
2. Blowing Air: When air is blown under the right-hand pan, the velocity of air flow under the pan ($v_{\text{below}}$) increases significantly compared to the static air above it ($v_{\text{above}}\approx 0$).
3. Pressure Drop: According to Bernoulli's equation, as the velocity under the pan increases, the pressure under the pan ($P_{\text{below}}$) must decrease. Since the air above the pan is not blown, the pressure above the pan remains at atmospheric pressure ($P_{\text{above}}=P_{\text{atm}}$).
Therefore, we have:
$P_{\text{above}}>P_{\text{below}}$

4. Net Force: This pressure difference creates a net downward pressure force on the right-hand pan. The net force ($F$) can be expressed as:
$F=(P_{\text{above}}-P_{\text{below}})\times A$
where $A$ is the surface area of the pan. Since $P_{\text{above}}>P_{\text{below}}$, the net force points vertically downwards.

As a result of this net downward force, the right-hand pan will move down.