Question

A thin metal disc of radius \(r\) floats on water surface and bends the surface downwards along the perimeter making an angle \(\theta\) with vertical edge of the disc. If the disc displaces a weight of water \(W\) and surface tension of water is \(T\), then the weight of metal disc is

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

Hint:

Always analyze the geometry of the "bend" to determine the direction of the surface tension force. Since the disc bends the surface downwards, the elastic nature of the water surface tries to pull it upwards to restore flatness.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For any object to float in equilibrium on the surface of a liquid, the sum of all downward forces must be exactly balanced by the sum of all upward forces.
In this scenario, the downward force is the weight of the metal disc itself (\( W_{disc} \)).
The upward forces are more complex:
1. The Buoyant Force (Archimedes' Upthrust): This is caused by the displaced volume of water and is equal to the weight of water displaced (\( W \)).
2. The Surface Tension Force: Because the disc is small and thin, surface tension plays a significant role. The "bending" of the surface indicates that the liquid pulls back on the disc.
Surface tension acts along the boundary line where the liquid, solid, and air meet (the perimeter).
Step 2: Key Formula or Approach:
The perimeter of the circular disc is \( L = 2\pi r \).
Surface tension \( T \) acts tangentially to the surface. Since the surface is bent downwards, the force on the disc has an upward vertical component.
Total upward force from surface tension \( F_T = T \times \text{Length} \times \cos(\text{angle with vertical}) \).
Given \( \theta \) is the angle with the vertical edge.
Step 3: Detailed Explanation:
The surface tension force acts along the entire circular boundary of the disc.
The length of this boundary is the circumference: \( 2\pi r \).
At every point on this circumference, the force vector has a magnitude \( T \) per unit length.
The problem states that the surface makes an angle \( \theta \) with the vertical edge of the disc.
The vertical component of this force per unit length is \( T \cos \theta \).
Therefore, the total upward force provided by surface tension is:
\[ F_{ST} = (2\pi r) \times (T \cos \theta) \]
Additionally, there is the standard buoyant force \( F_B \).
According to the question, the weight of water displaced is \( W \). By Archimedes' principle:
\[ F_B = W \]
Equating the downward force (weight of disc) to the sum of upward forces:
\[ W_{disc} = F_B + F_{ST} \]
Substitute the derived expressions:
\[ W_{disc} = W + (2\pi r)(T \cos \theta) \]
Rearranging to match the options:
\[ W_{disc} = 2\pi r T \cos \theta + W \]
This equilibrium ensures the disc remains on the surface. If the weight were higher, the disc would sink.
Step 4: Final Answer:
The weight of the metal disc is \( 2\pi r T \cos \theta + W \).
This is Option (C).