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

The correct option is 2πrTcosθ + W.

To find the weight of the metal disc, we need to analyze the forces acting on it when it is floating in equilibrium on the water surface.
Let us identify the vertical forces acting on the disc:
1. Weight of the metal disc ($W_d$): This force acts vertically downwards.
2. Buoyant force ($F_b$): The disc displaces a weight of water $W$. According to Archimedes' principle, the upward buoyant force is equal to the weight of the water displaced by the disc:
$F_b=W$
3. Force due to surface tension ($F_s$): The surface tension of water $T$ acts along the perimeter of the disc. The perimeter of the disc is:
$L=2\pi r$
The force due to surface tension acts at an angle $\theta$ with the vertical edge of the disc. Specifically, because the surface is bent downwards, the surface tension pull on the disc's edge has a vertically upward component.
The vertical component of the surface tension force per unit length is $T\cos \theta$.
Integrating this along the entire perimeter of length $2\pi r$ gives the total vertical upward force due to surface tension:
$F_{s,v}=T\left(2\pi r\right)\cos \theta =2\pi rT\cos \theta$

Since the disc is floating in equilibrium, the total downward force must balance the total upward force:
$\text{Downward Force}=\text{Upward Forces}$
$W_d=F_{s,v}+F_b$
Substituting the expressions we found:
$W_d=2\pi rT\cos \theta +W$

Therefore, the weight of the metal disc is $2\pi rT\cos \theta +W$.