NEET Important Chapter - Mechanical Properties of Fluids

Mechanical Properties of Fluids is an important section as far as NEET is concerned. In this chapter we covered all mechanical properties and characteristics  of fluid (liquid and gasses) and discussed how fluids are different from solids. 

Fluid mechanics are considered as two types. First is fluid statics in which fluid or liquid is stationary (not moving). So in this part we have studied pressure, variation of pressure with depth or height and buoyancy force.

Another part is fluid dynamics in which we have studied about the dynamic nature of fluid and in this section is covered continuity equation, bernoulli equation etc. 

The problems with continuity equations, Bernoulli equations are of frequently asked for the NEET exam.

In the later we also discussed the real fluid and their characteristics like viscosity, surface tension and excess pressure and also understand the important concept of surface energy. Let us see what topic and concept of mechanical properties of fluids are important for the NEET Exam with solved examples.

Mechanical Properties of Fluids  - Important Topics 

• Mechanical properties of fluids notes

• Pascal’s Law

• mechanical properties of fluids formulas

• Continuity Equation

• Bernoulli’s Principle

• mechanical properties of fluids solutions

• properties of fluid - Viscosity

• Stokes Law 

• Surface Energy

Important Concepts of Mechanical Properties of Fluids

List of Important Formulae

Solved Examples 

1. A 100 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Ans: 

Given mass of the girl = 100 kg

The diameter of the heel is d = 1 cm 

The radius of the heel is = d/2 = 0.005 m

Now find area of the heel = A = $\pi r^{2}$ 

A = $\pi (0.005)^{2}$ = $7.85 \times (10)^{-5}$

Due to heel the force on the floor would be F = mg = $100\times 9.8$ = 980 N

Pressure exerted by the heel on the floor by formula - 

$P = \dfrac{F}{A}$

$P = 12.48 \times (10)^{6}$ $\frac{N}{m^{2}}$

Trick - For determining pressure find the normal force acting on the surface then divide it by area of surface.

2. Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg $m^{-3}$. Determine the height of the wine column for normal atmospheric pressure?

Ans:  

We know that the density of mercury, 

$\rho_1  = 13.6 \times 10^{3} (kg \times m^{-3})$

Density of French wine

$\rho_2  = 984 (kg \times m^{-3})$

We know that height of the mercury column, $h_1$= 76 mm = 0.76 m

Let consider the height of the wine column = $h_2$

We know that Acceleration due to gravity, g = 9.8 $\frac{m}{s^{2}}$

Pressure in both column mercury and wine is equal so 

$\rho_1 \times h_1 \times g = \rho_2 \times h_2 \times g$

$h_2 = \dfrac{13.6 \times 10^{3} \times 0.76}{984} = 10.5$ 

Key Point: Since both the liquid are at same atmospheric pressure therefore  pressure remains the same for both the liquid.

Previous Year Questions from NEET Paper

1. A capillary tube of radius r is immersed in water and water rises in it to a height at the mass of the water in the capillary is 5 gram another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this cube is (NEET 2020)

Ans:

Given - radius of  capillary tube is  - r

Mass of the water in the capillary is - 5 gm 

New radius of capillary tube is - 2r

Change in pressure -

$\Delta P = \dfrac{2S}{R} = \rho gh$

Multiply the whole equation by $R^2$

$R^{2} \dfrac{2S}{R} = \rho gh R^{2}$

$2SR = \rho g V$

Here ρ is density of liquid (water), g is acceleration due to gravity, r is radius of tube and V is the volume.

By the formula of density 

M = ρV

$2SR = M g$

Therefore M is proportional to the radius of the tube. 

The mass of water that will rise in this cube is two times of 5 = 10 gm

Trick - In this type of question try to make the relation between given and required physical quantity.

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

Ans:

Given - Area of cross section = A = $mm^{2}$

The rate of flow = Q = Av

Velocity = v = $\sqrt{2 \times g\times h}$

=Q = $2 \times 10^{-6} \times \sqrt{2 \times 10\times2}$

=$2 \times 10^{-6}\times 2\times \sqrt{ \ 10}$

=$4 \times 10^{-6}\times 2\times 3.16$

=$12.64 \times 10^{-6}$ $\dfrac{m^3}{s}$

Trick - remember all the important formulas and see what is given and what we have to find then apply the relevant formula. In this question, think about the continuity equation and formula of rate of flow of fluid.

Practice Questions

1. Water falls down from a water tap with speed 3 m/s from 2 $m^{2}$ area then find area of streamline line when water goes down 80 cm (Ans: 1.2 $m^{2}$)

2. Pressure of water in a pipe when tap is open is $2 \times 10^{5} \frac{N}{m^{2}}$ and it become $2.5 \times 10^{5} \frac{N}{m^{2}}$ when it is closed then what is velocity of flow when tap was open? (Ans: 10 m/s)

Conclusion

In this article, we discussed important topics , important  Mechanical Properties of Fluid formulas from the NEET point of view. Students must make sure that they do not miss any of the above important topics to obtain a good score in the NEET exam.