Surface Tension

A metallic wire of density $d$ floats horizontally in water. The maximum radius of the wire such that it may not sink will be (surface tension of water $=T$) (Neglect buoyant force)

A wire is bend at an angle $\theta$. A rod of mass $m$ can slide along the bended wire without friction as shown in figure. If a soap film is maintained in the frame and frame is kept in a vertical position and rod is in equilibrium. If rod is displaced slightly in vertical direction. The time period of small oscillation is $a\pi \sqrt{\frac{bl}{g}}$ sec. where $a$ and $b$ are constant number then $a+b$ is

A water jet of radius $R$ is shown in the figure. The force between the part $1$ and part $2$ at the section $ABCD$ due to the surface tension is : (Assume that, $T$ is surface tension)

There is a uniform rectangular wire frame having a thin film of soap solution. A massless thin wire of radius $R$ and area of cross section $A$ is placed on the surface of film, and inside portion of the film is pricked. If surface tension of soap solution is $S$ and Young's modulus of wire is $Y$ then change in radius of the wire is:

A rectangular massless blade (as shown in figure) is placed on a water surface (surface tension $T=\frac{1}{7}\times {10}^{-1} \mathrm{N} {\mathrm{m}}^{-1}$. Perimeter of the middle opening is $4a$. Minimum force needed to lift this blade-up is $\left(a=5 \mathrm{cm}\right)$

A soap film is created in a small wire frame as shown in the figure. The sliding wire of mass $m$ is given a velocity $u$ to the right and assume that $u$ is small enough so that film does not break. Plane of the film is horizontal and surface tension is $T$. Then time to regain the original position of wire is equal to :

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $2.0\times {10}^{-2} \mathrm{N}$ (see figure). The length of the slider is $10 \mathrm{cm}$ and its weight negligible. The surface tension of the liquid film is :

The force required to drag a circular ring flat plate of radius $5 \mathrm{cm}$ on the surface of water is (ST of water is $75$ dyne/cm):

A cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the. soap bubble to reduce its size by half will be : (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)

A spherical soap bubble of radius $1.0 \mathrm{cm}$ is formed inside another of radius $2 \mathrm{cm}$. If a single soap bubble is formed which maintains the same pressure difference as inside the smaller and outside the larger bubble, the radius of this bubble is

A soap bubble of diameter $8 \mathrm{mm}$ is formed in air. The surface tension of liquid is $30$ $\mathrm{dyne} {\mathrm{cm}}^{-1}$. The excess pressure inside the soap bubble is :

A water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop as compared to smaller drops will be :

A thin metal ring of internal radius $8 \mathrm{cm}$ and external radius $9 \mathrm{cm}$ is supported horizontally from the pan of a balance so that it comes in contact with water in a glass vessel. It is found that an extra weight of $7.48 \mathrm{g}$ is required to pull the ring out of water. The surface tension of water is

A soap - bubble with a radius ' $r$ ' is placed on another bubble with a radius $R$ (figure). Angles between the films at the points of contact will be -

When charge is given to a soap bubble, it shows:

The shape of a liquid drop is spherical due to its property of :

A thin wire is bent in the form of a ring of diameter $3.0 \mathrm{cm}$. The ring is placed horizontally on the surface of soap solution and then raised up slowly. Upward force necessary to break the film formed between the ring and the solution is -

Three surfaces of liquids are shown here. Find the correct statements.

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

The force required to drag a circular flat plate of radius $5 \mathrm{cm}$ just touching the surface of water is (ST of water is $75$ dynelcm):