$A$ wire of length $20 \ cm$ is placed horizontally on the surface of water and is gently pulled up with a force of $1.456 \times 10^{-2} \ N$ to keep the wire in equilibrium. The surface tension of water is (in $N \ m^{-1}$)

$A$ drop of some liquid of volume $0.04 \ cm^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 \ cm^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5} \ dyne$ has to be applied normal to the surfaces. The surface tension of the liquid is (in $dyne \ cm^{-1}$):

The unit of surface tension in the $SI$ system is

The surface tension of a liquid is $70 \, dyne/cm$. In $MKS$ system its value is

The force necessary to pull a circular plate of $5\, cm$ radius from a water surface,for which the surface tension is $75\, dynes/cm$,is

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