• Written By Rumela_M
  • Last Modified 05-09-2022

Elasticity JEE Advanced Past Year Solved Questions in Steps

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Elasticity JEE Advanced Previous Year Questions with Solutions: Elasticity is one of the significant topics for JEE Advanced exam, with a weightage of around 2.42% (including Viscosity). Being an imperative topic, Elasticity requires good revision and practice. Candidates can go through previous years’ papers to keep track of Elastic topics frequently asked question and question weightage and increase the chances of securing more marks in the examination.

When force is applied to a material body, it gains a new length, shape and volume. However, when this force is removed, the body is able to return to its original shape and size. This property of the body to regain its original shape or size when external forces are removed is called elasticity. Continue reading the article to know the Elasticity JEE Advanced Past year solved questions.

Elasticity JEE Advanced Past Year Solved Questions

Elasticity is a chapter that is common to the JEE Main and JEE Advanced syllabuses. If candidates want to score good marks on the physics paper, then they must go through the solved questions and learn the techniques and tricks to solve the questions. It is very important for a candidate to learn these formulas and tricks to solve the questions in an exam-like environment where there is pressure and limited time.

For the feasibility of candidates who aspire to write either exam, we have provided the questions on elasticity that candidates can expect in their JEE Main or Advanced question paper:

Elasticity JEE Advanced Past Year Solved Questions With Solutions

Here are some of the Elasticity questions asked in the previous question paper of JEE Advanced:

Q.1: An elastic string of length 42 cm and a cross-section area of 10–4 m2 is attached between two pegs at a distance of 6mm, as shown in the figure. A particle of mass m is kept at the midpoint of string stretched as shown in the figure by 20 cm and released. As the string attains natural length, the particle attains a speed of 20 m/s. Then young modulus Y of the string is of order:

(A) 108
(B)1012
(C)106
(D) 104

Answer:(C)

Solution:

Elastic stream energy × volume = kinetic energy

Y = [(0.05 x400 x0.42)/(0.2)2 x 10-4]
Y = 2.1 x 106 N/m2

Q.2: One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:

(A) 0.25
(B) 0.50
(C) 2.00
(D) 4.00

Answer: (C) 2.00

Solution:

For thick wire,

For thin wire,

Δl2/Δl1 = 2

Q.3: A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m, and its cross-sectional area is 4.9 x 10-7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs a simple harmonic motion of angular frequency 140 rad s–1. If Young’s modulus of the material of the wire is n x 109 Nm-2, the value of n is?

(A) 4
(B) 2
(C) 8
(D) 1

Answer: (A) 4

Solution:

The frequency of the oscillation can be written as
ω = √(YA/mL)
⇒ Y = ω2mL/A

Elasticity Class 11 Previous Year Questions – JEE Main

Here are some questions on Elasticity that have been previously asked in the JEE Main examinations:

Q1: A compressive force, F, is applied at the two ends of a long thin steel rod. It is heated simultaneously such that its temperature increases by ΔT. The net change in its length is zero. Let L be the length of the rod. A is its area of cross-section. Y is Young’s modulus, and α is its coefficient of linear expansion. Then, F is equal to:

(a) L 2YαΔT
(b) AY/αΔT
(c) AYαΔT
(d) LAYαΔT

Solution: Thermal expansion, ΔL = LαΔT ——-(1) Compression ΔL’ produced by applied force is given by, Y = FL/AΔL’ ⇒ F = YAΔL’ /L————-(2) Net change in length = 0 ⇒ΔL’ = ΔL ——(3) From (1),(2) and (3) F = YA x (LαΔT)L = YAαΔT

Answer: (c) AYαΔT

Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is:
(a) 0.2 J
(b) 10 J
(c) 20 J
(d) 0.1 J

Solution: Elastic energy per unit volume = ½ x stress x strain Elastic Energy = ½ x stress x strain x volume = ½ x F/A x (ΔL /L) x (AL) = ½ x FΔL = ½ x 200 x 10-3 Elastic Energy = 0.1 J

Answer: (d) 0.1 J

Q3: A rod of length L at room temperature and uniform cross-section A area is made of metal with a coefficient of linear expansion α. It is observed that an external compressive force F is applied to each of its ends, prevents any change in the length of the rod when its temperature rises by ΔT K. Young’s modulus, Y for this metal is:
(a) F/AαΔT
(b) F/Aα(ΔT – 273)
(c) F/2AαΔT
(d) 2F/AαΔT

Solution: Young’s Modulus Y = stress/strain = F/A(Δl/l) Substituting the coefficient of linear expansion α =Δl /(lΔT) Δl /l= αΔT Y= F/A(αΔT)

Answer: (a) F/AαΔT

Q4: Young’s moduli of two wires, A and B, are in the ratio of 7:4. Wire A is 2m long and has a radius R. Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a given load, then the value of R is close to:
(a) 1.5 mm
(b) 1.9 mm
(c) 1.7 mm
(d) 1.3 mm

Solution: Δ1= Δ2 (Fl1/πr1 2y1) = (Fl2/πr2 2y2) 2/(R2 x 7)= 1.5/(22 x 4) R= 1.75 mm

Answer: (c) 1.7 mmC

Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?
(a) 1 mm
(b) 1.15 mm
(c) 0.90 mm
(d) 1.36 mm

Solution

Stress = F/A
Stress = 400 x 4/πd2
= 379 x 106 N/m2
d2 = (400 x 4)/(379 x 106π)
d = 1.15 mm.

Elasticity JEE Advanced Questions – Tips and Tricks To Solve Questions

Here are some tips and tricks to solve the Elasticity questions for JEE Advanced question paper efficiently:

  • The candidate can start working on the questions by making a diagram. 
  • A diagram helps the candidate to visualise properly the stress being applied on mass and other variables mentioned in the question.
  • Candidates should keep the formulas of the dimensions in mind while trying to solve their questions. This will prevent them from getting confused with equations.

FAQs on Elasticity JEE Advanced Previous Year Questions With Solutions

The following are the most frequently asked questions on Elasticity JEE Advanced previous year questions with solutions:

Q.1: When will the JEE Advanced 2023 be conducted?

Ans: The JEE Advanced 2023 exam dates will be released soon.

Q.2: Who will release the exam pattern for JEE Advanced 2023?

Ans: IIT Bombay will release the exam pattern for JEE Advanced exam 2023. 

Q.3: What is the duration of the JEE Advanced examination?

Ans:  Each paper will be conducted for 3 hours. PwD candidates will get 4 hours to attend each paper. 

Q.4: What is elasticity?

Ans: The elasticity of an object is the property by which a body tends to return to its original shape, size, or form when the external forces that have been applied to it are removed.

Q.5: Where can I practice mock test questions on JEE Advanced 2023?

Ans: Candidates can practice free mock test questions related to JEE Advanced examinations on the Embibe app.

Important links:

JEE Advanced Mock TestIIT JEE Advanced – All Guides
JEE Advanced TestJEE Advanced Preparation

We hope the above article has answered all your doubts and queries. Embibe wishes you good luck and all the best for JEE Advanced 2023.

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