Question

A thin metal disc of radius $r$ floats on a water surface and bends the surface downwards along the perimeter making an angle $\theta$ with the vertical edge of the disc. If the weight of water displaced by the disc is $W$, the weight of the metal disc is [$T$ = surface tension of water]

• A. $2\pi r T \cos\theta + W$
• B. $W - 2\pi r T \cos\theta$
• C. $2\pi r T + W$
• D. $2\pi r T \cos\theta - W$

Hint:

Remember that surface tension acts like an elastic trampoline web. When the disc pushes down, the web pulls upward along the edge to help support the weight. Therefore, the surface tension force must add to the buoyant force ($W$) to support the heavier object, eliminating options (B) and (D).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem describes a state of static equilibrium for a thin circular metal disc floating on the surface of water.
The weight of the disc pulls it downward, causing the water surface to bend along its perimeter. This deflection sets up a surface tension boundary force at an angle $\theta$ relative to the vertical edge.
The liquid also provides an upward buoyant force equal to the weight of the displaced fluid ($W$). We need to determine the total weight of the disc.

Step 2: Key Formula or Approach:
For the floating disc to remain completely stable, the downward forces must balance the upward forces:
$$\sum F_{upward} = \sum F_{downward}$$ The downward force is simply the total gravitational weight of the disc ($W_{disc}$).
The upward forces supporting the disc consist of two distinct components:
1. The vertical component of the force due to surface tension ($F_{st}$) acting all along the perimeter loop.
2. The buoyant upthrust force ($F_b$), which equals the weight of the displaced water ($W$).

Step 3: Detailed Explanation:
Surface tension force acts along the contact perimeter of the disc. The perimeter length of a circle of radius $r$ is $2\pi r$.
The total surface tension force magnitude is $F_{total} = T \times (2\pi r)$.
The problem states that the liquid surface meets the vertical edge of the disc at an angle $\theta$. The upward vertical component of this surface tension force is resolved using the cosine component:
$$F_{st, y} = T(2\pi r)\cos\theta = 2\pi r T \cos\theta$$ The buoyant upthrust supporting the system is equal to the weight of the displaced liquid:
$$F_b = W$$ Equating the forces for equilibrium:
$$W_{disc} = F_{st, y} + F_b$$ $$W_{disc} = 2\pi r T \cos\theta + W$$

Step 4: Final Answer:
The total weight of the metal disc is given by $2\pi r T \cos\theta + W$, which corresponds to option (A).