Since the soap bubble floats in the air, the sum of the mass of the wall and the enclosed helium gas should be equal to the mass of the displaced air. The density of helium gas is less than half of the density of the air. The mass of the wall should be greater than that of the enclosed gas.

Our sensation of taste is via tongue. the more the tongue tingles, the more we feel tasty. Cold soup has more surface tension because of low temperature resulting high attraction between the molecules. on the other hand, hot soup has low surface tension because of which, instead of getting attracted to other soup molecules,it sticks to our tongue and it tingles. thus, hot soup feels tastier than cold soup.

pascal's law states that, for a liquid enclosed in a vessel , pressure at all the portions of the container is equal and when an external pressure is applied, it transmits equally in all directions.

as stated by pascal's law, for an external force applied , the pressure is distributed within , the distribution is only dependent on area of pistons instead of the vessel.

for a cylindrical vessel , the height should be equal to its diameter for making the force equal in wall and bottom.

According to Pascal’s Law,“The external static pressure applied on a confined liquid is distributed or transmitted evenly throughout the liquid in all directions”.

Calculation of fluid pressure

• Pfluid= P + ρgh
• where,
• P= Pressure at the reference point
• Pfluid= Pressure at a point in a fluid
• ρ= Density of the fluid
• g= Acceleration due to gravity (considering earth g = 9.8 m/s)
• h= Height from the reference point
• The density of a fluid may be estimated by dividing its mass by the volume of fluid...

law of flotation, for a body of density D and liquid L

D<L , body floats

D>L , body sinks

D=L, equilibrium.

yes, for D=L , the body is in equilibrium

Let$M$be the mass of wooden cube and$a$be the side of the cube.

The body is in equilibrium in both cases. So lets equate the forces in both cases.
Case I:
$0.2g+Mg={\rho }_w{V}_{1}g......\left(1\right)$

Case II:
$Mg={\rho }_w{V}_{2}g.......\left(2\right)$

Here,$V_1=V_2+a^2.\left(0.02\right)V_1{V}_2=a^2.\left(0.02\right)$

Now, by subtracting$\left(2\right)$from$\left(1\right)$, we get
$0.2g+MgMg={\rho }_w\left(V_1{V}_2\right)g$
$0.2g={\rho }_{w}g\left(\mathrm{0.02.}{a}^2\right)$
$a^2=\frac{0.2}{0.02{\rho }_w}=\frac{10}{1000}=0.01m^{2}a=0.1ma=10cm$

Streamlineflowor a laminar flow is an uninterrupted flow (as of air) past a solid body in which the direction at every point remains unchanged with the passage of time.

Here, at low fluid velocities, there are no turbulent velocity fluctuations and the fluid tends to flow without lateral mixing. Here, the motion of particles of the fluid follows a particular order with respect to the particles moving in a straight line parallel to the wall of the pipe such that the adjacent layers slide past...

Viscosity is a measure of a fluid’s resistance to flow.

The viscosity of liquids decreases rapidly with an increase in temperature, and the viscosity of gases increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas gases flow more slowly. Also, viscosity does not change as the amount of matter changes, therefore it is an intensive property.

Solution ,

Given ,

Density of water (ρ1) = 1000 kg/m3

Density of glycerin (ρ2) = 1200 kg/m3

Height of glycerin (h2) = 4 cm

The height difference of both columns of the pipe. = ?

we have ,

The height of the column of the pipe (h1) :

ρ1h1= ρ2h2

(1000)(h1) = (1200)(4)

(1000)(h1) = 4800

h1= 4.8 cm

The height difference of both columns of the pipe U = h1– h2= 4.8 cm – 4 cm = 0.8 cm