Surface Tension

A large tank is filled with two liquids of specific gravities $\text{2}\text{σ}$ and $\text{σ}$. Two holes are made on the wall of the tank as shown. Find the ratio of the distances from $O$ of the points on the ground where the jets from holes $A \mathrm{and} B$ strike.​

Water rises to a height of $10 \mathrm{cm}$ in a capillary tube and mercury falls to a depth of $3.5 \mathrm{cm}$ in the same capillary tube. If the density of mercury is $13.6 \mathrm{gm} {\mathrm{cc}}^{-1}$ and its angle of contact is $135°$ and density of water is $1 \mathrm{gm} {\mathrm{cc}}^{-1}$ and its angle of contact is $0°$, then the ratio of surface tensions of the two liquids is $(\cos 135°=0.7)$

Two arms of a U-tube have unequal diameters $d_1=1.0 \mathrm{mm}$ and $d_2=1.0 \mathrm{cm}$. If water (surface tension $7\times {10}^{-2} \mathrm{N}/\mathrm{m}$) is poured into the tube held in the vertical position, find the difference in the level of water in the U-tube. Assume the angle of contact to be zero.

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. If the atmospheric pressure is H m of mercury & the density of mercury is n times that of lake water. Find the depth of the lake

On dipping one end of a capillary in liquid and inclining the capillary at an angles 30^o and 60^​o with the vertical, the lengths of liquid columns in it are found to be $l_1$ and $l_2$ respectively. The ratio of $l_1$ and $l_2$ is

On dipping one end of a capillary in liquid and inclining the capillary at an angles ${30}^0$ and ${60}^0$ with the vertical, the lengths of liquid columns in it are found to be $l_1$ and $l_2$ respectively. The ratio of $l_1$ and $l_2$ is

The radii of two soap bubbles are $r_1$ and $r_2$ . In isothermal conditions, two meet together in vacuum. Then the radius of the resultant bubble is given by

Water rises in a vertical capillary tube upto a height of $2.0 \mathrm{cm}$. If the tube is inclined at an angle of $60°$with the vertical, then upto what length the water will rise in the tube

Water rises in a vertical capillary tube up to a height of $2.0 \mathrm{cm}$. If the tube is inclined at an angle of $60°$ with the vertical, then up to what length the water will rise in the tube?

Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is

The excess pressure inside an air bubble of radius r just below the surface of water is ​$P_1$ . The pressure inside a drop of the same radius just outside the surface is $P_2$ . If T is surface tension then

If the radius of a soap bubble is four times that of another,then the ratio of their excess pressures will be

A film of water is formed between two straight parallel wires of length $10 \mathrm{cm}$ each separated by $0.5 \mathrm{cm}$. if their separation is increased by $1 \mathrm{mm}$ while still maintaining their parallelism, how much work will have to be done (surface tension of water $= 7.2\times {10}^{-2} \mathrm{N} {\mathrm{m}}^{-1}$)

A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy.( Surface tension of mercury is 0.465 J/m² )

Radius of a soap bubble is increased from $R$ to $2R$  work done in this process in terms of surface tension is

Radius of a soap bubble is increased from $R$ to $2R$ work done in this process in terms of surface tension is

A liquid film is formed over a frame $ABCD$ as shown in figure. Wire $CD$ can slide without friction. The mass to be hung from $CD$ to keep it in equilibrium is

There is horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius $R$. If the surface tension of the loop be $T$ , then what will be the tension in the thread​

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

A wooden stick $2 \mathrm{m}$ long is floating on the surface of water. The surface tension of water $0.07 \mathrm{N} {\mathrm{m}}^{-1}$. By putting soap solution on one side of the stick, the surface tension is reduced to $0.06 \mathrm{N} {\mathrm{m}}^{-1}$. The net force on the stick will be