Force due to Surface Tension

A square frame of each side L is dipped in a soap and taken out, the force acting on the film formed is ( T = surface tension of soap solution)

A metal wire of length '$L$' and density '$d$' rests horizontally on the free surface of water. The maximum radius of the wire so that the wire does not sink in water is

[$T$ = surface tension of water, $g$ = gravitational acceleration]

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

What is the radius of the biggest aluminium coin of thickness $t$ and density $\rho$, which will be able to float on the water surface of surface tension $T$? ($g$ = acceleration due to gravity)

A metal wire of length 'L' and density 'd' floats horizontally on the free surface of water. The maximum radius of the wire does not sink in water is

[ $T$ = surface tension of water, $g$ = gravitational acceleration]

A disc of paper of radius ' $R$ ' has a hole of radius ' $r$ '. It is floating on a liquid of surface tension ' T '. The force of surface tension on the disc is

A liquid drop of volume V is placed on the surface of glass plate. Then another glass plate is placed on it such that liquid forms a thin layer of area A between the surfaces of two plates. To separate the plates a force F has to be applied normal to the surfaces. The surface tension of the liquid is

A water drop of $0.01\text{ cm}^{3}$ is squeezed between two glass plates and spreads in to area of $10\text{ cm}^{2}$. If surface tension of water is 70 dyne $/\text{cm}$ then the normal force required to separate glass plates from each other will be

A thin flat circular disc of radius $4.5\text{ cm}$ is placed gently over the surface of water. If surface tension of water is $0.07\text{ N m}^{-1}$, then the excess force required to take it away from the surface is

Liquid drops are falling slowly one by one from vertical glass tube. The relation between the weight of a drop ' $w$ ', the surface tension ' $T$ ' and the radius ' $r$ ' of the bore of the tube is (Angle of contact is zero)

A liquid drop of density ' $\rho$ ' is floating half immersed in a liquid of density ' $d$ '. If ' $T$ ' is the surface tension then the diameter of the liquid drop is ( $g =$ acceleration due to gravity)

A steel coin of thickness '$d$' and density '$\rho$' is floating on water of surface tension '$T$'. The radius of the coin $(R)$ is [$g =$ acceleration due to gravity]

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5 \times 10^{-4}$ Ncm${}^{-1}$. The radius of drop in $\text{cm}$ will be:

(g = 10 ms${}^{-2}$)

The force required to take away a flat circular plate of radius $2\text{ cm}$ from the surface of water is $[$Surface tension of water $= 70 \times 10^{-3}\text{ N m}^{-1}, \pi = \frac{22}{7}]$

A needle is $7\text{ cm}$ long. Assuming that the needle is not wetted by water, what is the weight of the needle, so that it floats on water?

$[T =$ surface tension of water $= 70\,\frac{\text{dyne}}{\text{cm}}]$

[acceleration due to gravity $= 980\text{ cm s}^{-2}$]

A thin metal disc of radius 'r' floats on 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 drop of liquid of density '$\rho$' is floating half immersed in a liquid of density '$d$'. If '$T$' is the surface tension, then the diameter of the drop of the liquid is

A square frame of each side $L$ is dipped in a soap and taken out. The force acting on the film formed is ($T =$ surface tension of soap solution)

A small spherical droplet of density d is floating exactly half immersed in a liquid of density $\rho$ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms cylindrical surface on the side that gives rise to lower pressure in the water in comparison to atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then the pressure in water between the plates is lower by:

On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius $R$ and making a circular contact of radius $r$ with the bottom of the vessel. If $r \ll R$ and the surface tension of water is $T,$ value of $r$ just before bubbles detach is: (density of water is $\rho_w$)

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times 10^{-2}\text{ N}$ (see figure). The length of the slider is $30\text{ cm}$ and its weight negligible. The surface tension of the liquid film is