 Most Affordable JEE | NEET | 8,9,10 Preparation by Kota's Top IITian Doctor Faculties

# Elasticity - JEE Advanced Previous Year Questions with Solutions JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.     Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions    Download eSaral app  for free study material and video tutorials.   Simulator   Previous Years JEE Advanced Questions
Q. 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 $4.9 \times 10^{-7} \mathrm{m}^{2}$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s–1. If the Young’s modulus of the material of the wire is $\mathrm{n} \times 10^{9} \mathrm{Nm}^{-2}$, the value of n is ? [IIT-JEE 2010]
Ans. 4 $\mathrm{F}=6 \mathrm{A}=\mathrm{YSA}=\frac{\mathrm{YA}}{\ell} \cdot \mathrm{x}$ $140=\sqrt{\frac{\mathrm{n} \times 10^{9} \times 4.9 \times 10^{-7}}{1 \times 0.1}}=70 \times \sqrt{\mathrm{n}}$ $\therefore \mathrm{n}=4$
Q. 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 [JEE-Advance-2013]
Ans. (C) Force same = F $\mathrm{Y}=\frac{F / A}{\Delta L / L}$ $F=Y A_{1} \frac{\Delta L_{1}}{L_{1}}=Y A_{2} \frac{\Delta L_{2}}{L_{2}}$ $\frac{\Delta L_{2}}{\Delta L_{1}}=\frac{L_{2}}{L_{1}} \times \frac{A_{1}}{A_{2}}$
Q. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are):- (A) P has more tensile strength than Q (B) P is more ductile than Q (C) P is more brittle than Q (D) The Young's modulus of P is more than that of Q [JEE-Advance-2015]
Ans. (A,B) Slope of this graph represents the reciprocal of Young's modulus. since Slope of P > Slope of Q Hence Y of P < Y of Q

Sahil
June 1, 2023, 7:04 a.m.
Figure missing in third question
Apparantly a genius because these ques are very easy
Dec. 19, 2021, 8:53 a.m.
Easy as hell wtf bro is this the type of ques in jee advance you've got to be kidding me
Aashrith
Dec. 10, 2020, 8:46 p.m.
only 3 qn??
munna bhai
Nov. 10, 2020, 2:48 p.m.
PRETTY SIMPLE QUESTIONS. I WANT NEW QUESTIONS TO PREPARE PLEASE KEEP IN THIS WEBSITE ONLY SIR
Harshit jain
Nov. 2, 2023, 6:35 a.m.
Yes
Abcdefghijklmnopqrstuvwxyz
June 18, 2020, 10:52 p.m.
Why only these  