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Last updated date: 26th May 2023

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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.

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

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.

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.

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)

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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.

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.

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 %)

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)

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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JEE Main 2023 January and April Session exam dates and revised schedule have been announced by the NTA. JEE Main 2023 January and April Session will now be conducted on 24-Jan-2023 to 31-Jan-2023 and 6-Apr-2023 to 12-Apr-2023, and the exam registration closes on 12-Jan-2023 and Apr-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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JEE Main 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2023 with Vedantu right on time.

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Download JEE Main Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.

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In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.

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JEE Main 2023 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in IIT colleges. JEE Main 2023 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2023 state rank list on the official website or on our site.

Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.

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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.

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