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 ${}^{\prime }{W}^{\prime }$ and surface tension of water is '$T$', then the weight of metal disc is :

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

Answer 1

The correct option is C $2\pi rT\cos \theta +W$

The FBD as shown has three forces acting on the disc:

• weight mg of the metal disc downward
  
• force due to surface tension(T) upward at an angle $\theta$ with the vertical
  
• buoyancy force upward

Equating all the forces,
$mg=2\pi rT\cos \theta +W$
Note that the horizontal component of the surface tension force $T\sin \theta$ will cancel by symmetry