JEE Important Chapter - Properties of Solids and Liquids

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Properties of Solids and Liquids for JEE

The chapter Properties of solids and liquids is one of the most important chapters for JEE Main. It mainly deals with the behaviour of solids under stress and liquids at rest and motion.

Related to the properties of solid, we learn about types of stresses and strain followed by the types of  modulus of elasticity. The stress-strain curve of a material and Hooke's law is explained in detail in this chapter. Thermal stress and work done in stretching a wire are also discussed in this chapter in detail.

In properties of liquid, Pascal’s law and its applications are studied along with Archimedes's principles and numerical problems related to it. Bernoulli's theorem and its application and problems are very important topics in this chapter. Viscosity and surface tension is also explained in detail in this chapter. We have seen some conversion of solid to liquid examples and also liquid to solid in lower grades. A solid to liquid example of its conversion is the melting of ice.

Let us see what are solids and liquids properties and the important concepts and formulas needed for JEE Main in this chapter properties of solids and liquids with solved examples.

Properties of Solids and Liquids - Important Topics

• Elasticity of solids and its elastic behaviour

• Young’s modulus

• Bulk modulus

• Rigidity modulus

• Stress-strain curve

• Pascal’s Law

• Mercurial Barometer

• Archimedes Principle

• Bernoulli’s Therorem

• Application of Bernoulli’s theorem

• Stoke’s Law

• Surface tension and surface energy

Solved Examples

1. When a load of 5 kg is hung on a wire, then the extension of 3 m takes place. The work done will be,

Ans:

The mass of the load hung on the wire, m = 5 kg

The force acting on the wire is calculated as ,

$F=mg$

$F=5\times9.8$

$F=49~N$

We can calculate the work done in extending the wire by using the formula given below,

$W=\dfrac{1}{2}F\Delta l$

$W=\dfrac{1}{2}\times 49\times 3$

$W=73.5~J$

Key Point: The work done in extending the wire is half the product of the force acting on the wire and the extension of the wire.

1. In a capillary tube, water rises by 1.2 mm. The height of water will rise in another capillary tube having half the radius of the first will be ____.

Ans:

The formula to calculate the capillary rise of first tube is given by,

$h_1=\dfrac{2S\cos\theta}{\rho rg}$

$1.2~mm=\dfrac{2S\cos\theta}{\rho rg}$...(1)

The capillary rise of the water in the second tube is given by,

$h_2=\dfrac{2S\cos\theta}{\rho (r/2)g}$

$h_2=\dfrac{4S\cos\theta}{\rho rg}$.....(2)

Divide the equation (2) by equation (1) to calculate the height of capillary rise in second tube.

$\dfrac{h_2}{1.2~mm}=\dfrac{\dfrac{4S\cos\theta}{\rho rg}}{\dfrac{2S\cos\theta}{\rho rg}}$

$h_2=2\times2~mm$

$h_2=4~mm$

Key Point:  The surface tension and the angle of contact in both tubes are equal since the same liquid is used in both tubes.

Previous Year Questions from JEE Exam

1. A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift _______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ‘m’ on the smaller piston. (JEE 2021)

(Image will be uploaded soon)

Ans:

Applying Pascal’s law, the pressure acting on smaller piston due to mass m is equal to the pressure acting on the larger piston.

Let A1 and A2 be the area of the smaller piston and larger piston respectively.

$\dfrac{mg}{A_1}=\dfrac{100g}{A_2}$

$\dfrac{A_1}{A_2}=\dfrac{m}{100}$...(1)

Now, let M be the mass that hydraulic press can lift after changing the area of cross section of both pistons.

Again applying pascal’s law,

$\dfrac{mg}{(A_1/16)}=\dfrac{Mg}{(16A_2)}$

$\dfrac{A_1}{A_2}=256\dfrac{m}{M}$...(2)

Using  equation (1) and equation (2), we can calculate the value of M.

$\dfrac{m}{100}=256\dfrac{m}{M}$

$M=25600~kg$

Therefore, the new mass it can lift is 25600 kg

Trick:  The pressure acting on both the piston are equal according to the Pascal’s law.

1. A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ________ cm. (JEE 2021)

Ans:

(Image will be uploaded soon)

The formula to calculate the young’s modulus of the wire is given by,

$Y=\dfrac{Fl}{A\Delta l_1}$....(1)

When the diameter is doubled, the area of cross section of the wire becomes four times initial area of cross section.

When the same force is applied after the length and diameter is doubled, then formula for young’s modulus is given by,

$Y=\dfrac{F(2l)}{(4A)\Delta l_2}$

$Y=\dfrac{Fl}{2A\Delta l_2}$....(2)

Using equation (1) and (2),

$\dfrac{Fl}{A\Delta l_1}=\dfrac{Fl}{2A\Delta l_2}$

$\Delta l_2=\dfrac{\Delta l_1}{2}$

$\Delta l_2=\dfrac{0.04}{2}$

$\Delta l_2=0.02~m=2~cm$

Therefore, the new extension will be 2 cm.

Trick: Young’s modulus deos not depend on the dimensions on the wire and only depends onthe material.

Practice Questions

1. An iceberg of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of the volume of ice cube outside the water is ? (Ans: 10 %)

2. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof will be (Ans: 2.4✕105 N)

Conclusion

In this article, we discussed important topics and formulas related to Properties of solids and liquids from the JEE 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 JEE Main exam.

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FAQs on JEE Important Chapter - Properties of Solids and Liquids

FAQ

1. How many questions are asked from properties os solids and liquids in JEE?

About 2-3 questions from this chapter are asked in JEE exam every year which corresponds to around 8-12 marks in the JEE Main exam.

2. Is the chapter properties of solids and liquids a tough chapter?

As far as teh JEE main exam is concerned, properties of solids and liquids is a moderate difficult chapter. But it is a vast chapte which consists of more theories and concepts and related formulas. But with proper learning and practices, all the concepts in this chapter can be learned without any difficulty.

3. How to get a good score in JEE Main to get into the best NITs?

To get into best NITs in the nation, we have to crack the JEE Main exam with good scores. Even though JEE Main is a very competitive exam, we can score good marks if we understand the concepts in depth and pratice lots of questions and finally doing all the previous year question papers without any fail.