If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

A U-tube, with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level. The density of the oil is

A small sphere of radius $r$ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to

A small hole of area of cross-section $2$ mm$^2$ is present near the bottom of a fully filled open tank of height $2$ m$.$ Taking $g = 10$ m/s$^2,$ the rate of flow of water through the open hole would be nearly:

A soap bubble, having radius of $1$ mm, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2}$ N/m. The pressure inside the bubble equals at a point $Z_0$ below the free surface of water in a container. Taking $g = 10$ m/s$^2,$ density of water $= 10^3$ kg/m$^3,$ the value of $Z_0$ is:

A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h.$ The mass of the water in the capillary is $5$ g. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:

The velocity of a small ball of mass $M$ and density $d,$ when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is $d_2,$ then the viscous force acting on the ball will be:

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball $(v)$ as a function of time $(t)$ is:

If a soap bubble expands, the pressure inside the bubble:

The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly : (surface tension of soap solution = 0.03 N m$^{-1}$)

The venturi-meter works on:

A thin flat circular disc of radius $4.5m$ is placed gently over the surface of water. If surface tension of water is $0.07$ N/m$,$ then the excess force required to take it away from the surface is: