NEET Exam  >  NEET Notes  >  Physics Class 11  >  NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids

NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids

2024

Q1: The maximum elongation of a steel wire of  1m length if the elastic limit of steel and its Young's modulus, respectively, are  8 × 10Nm-2 and 2 × 1011 Nm-2, is:             [2024]
(a) 4 mm
(b) 0.4 mm
(c) 40 mm
(d) 8 mm
Ans: 
(a)
First, we need to find the maximum force that can be applied to the steel wire within its elastic limit. This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.

Let's assume the area of cross-section of the wire is A. The force exerted can be given by:
2024
where:
σ=8×108N /m2 (Elastic limit of steel)
To calculate the elongation (ΔL) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:

2024
where:
Y = 2 x 1011 N / m2 (Young's modulus of steel),
L = 1 m (original length of the wire).
Substituting for F from the earlier expression and rearranging the formula, we get:

2024
Thus, the maximum elongation of the wire within the elastic limit is  4mm.
The answer is: Option A - 4 mm.

Q2: A metallic bar of Young's modulus,   0.5 × 1011 Nm-2 and coefficient of linear thermal expansion 10-5∘C-1, length  1 m and area of cross-section 10-3 m2 is heated from 0C to 100C without expansion or bending. The compressive force developed in it is :
(a) 5 x 103 N
(b) 50 x 103 N
(c) 100 x 103 N
(d) 2 x 103 N             [2024]
Ans:
(b)
Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating. Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (α), change in temperature (ΔT), and the original dimensions of the bar.
First, compute the linear expansion of the bar if it were free to expand. The change in length (ΔL) due to thermal expansion can be computed through the formula:
2024
Given:

2024
Thus:
2024 m
This is the change in length that the bar would undergo if not constrained.
However, in this scenario, the bar is constrained and does not actually expand. This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain
2024
Where the strain (ε) under constrained conditions due to thermal expansion is:
2024
Therefore:
2024
This stress is the force per unit area. To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar:
2024
Thus, the compressive force developed in the bar is  50×103 N.
Hence, the correct answer is Option B: 50×103 N.

Q3: A thin flat circular disc of radius 4.5cm is placed gently over the surface of water. If surface tension of water is 0.07Nm-1, then the excess force required to take it away from the surface is
(a) 19.8 mN
(b) 198 N
(c) 1.98 mN
(d) 99 N
Ans: 
(a)
2024

2024

2023

Q1: Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is       [2023]
A: 2W/A
B: W/A
C: W/2A
D: Zero
Ans: 
B
2023

2022

Q1: Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).           
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. 
Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below       [2022]
A: Both (A) and (R) are true and (R) is not the correct explanation of (A) 
B: (A) is true but (R) is false 
C: (A) is false but (R) is true 
D: Both (A) and (R) are true and (R) is the correct explanation of (A)
Ans: B
It is true that the stretching of spring is determined by shear modulus of the spring as when coil spring is stretched neither its length nor its volume changes, there is only change in its shape. The tensile strength of steel is more than that of copper. Hence Assertion is true and reason is false. 

2020

Q1: Dimensions of stress are :     [2020]
A: [ML0T-2
B: [ML-1T-2
C: [MLT-2
D: [ML2T-2]
Ans: 
B
2020

Q2: A wire of length L, area of cross-section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young's modulus is : [2020]
A:
2020
B:
2020
C:
2020
D:

2020
Ans: B
from young's modulus formula
2020

2019

Q1: When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:    [2019]
A: Mgl
B: MgL
C: 2019
D: 2019
Ans:
C
Solution:

Given:

  • Original length of the wire L.
  • Extended length L + lL+l, so the extension is ll.
  • Mass of the block M, so the force exerted by the block due to gravity is F = MgF=Mg, where gg is the acceleration due to gravity.

Step 1: Determine the Force Acting on the Wire

The force acting on the wire due to the suspended block is equal to the weight of the block:Step 1: Determine the Force Acting on the Wire

Step 2: Calculate the Elastic Potential Energy

The elastic potential energy U stored in a stretched wire (or spring) with an extension ll is given by:Step 2: Calculate the Elastic Potential Energy

2018

Q1: Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount ?    [2018]
A: 9F
B: 6F
C: 4F
D: F
Ans:
A
Solution:
2018

2017

Q1: The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', the fractional decrease in radius is:-    [2017]
A: 2017
B: 2017
C: 2017
D: 2017
Ans:
C
Solution:

2017

2014

Q1: Copper of fixed volume 'V' is drawn into wire of length 'l'. When this wire is subjected to a constant force 'F', the extension produced in the wire is 'Δl'. Which of the following graphs is straight line?    [2014]
A: Δl versus l/l2
B: Δl versus l
C: Δl versus l/l
D: Δl versus l2
Ans: 
D
Solution:
2014

There is no Question for NEET 2021

The document NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids is a part of the NEET Course Physics Class 11.
All you need of NEET at this link: NEET

FAQs on NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids

1. What are the main types of stress and strain that appear in NEET physics questions on mechanical properties?
Ans. Stress measures force per unit area applied to a material, while strain quantifies the deformation produced. NEET previous year questions focus on three stress types: tensile (pulling), compressive (pushing), and shear stress. Corresponding strains are longitudinal, volumetric, and shear strain. Understanding these distinctions is critical for solving numerical problems on elastic behaviour and material deformation in Class 11 physics.
2. How do I calculate Young's modulus and identify when to use it in NEET exam questions?
Ans. Young's modulus equals stress divided by strain for materials undergoing longitudinal deformation. Apply it when questions describe stretching or compressing solids along their length. NEET previous year papers frequently test this concept through numerical problems involving wire extension, elastic constants, and material properties. The formula E = (F/A)/(ΔL/L) appears repeatedly across 2014-2025 question sets.
3. Why do some materials show elastic behaviour while others become permanently deformed in mechanical properties problems?
Ans. Materials exhibiting elastic behaviour regain original shape after stress removal because internal forces restore molecular positions. Permanent deformation occurs when stress exceeds the elastic limit, causing plastic deformation where atoms shift into new positions permanently. NEET questions test this distinction through scenarios involving wire stretching, spring compression, and material breaking points, requiring students to identify elastic vs. plastic regions.
4. What's the difference between bulk modulus and shear modulus that keeps appearing in NEET previous year questions?
Ans. Bulk modulus measures resistance to volume change under uniform pressure, while shear modulus measures resistance to shape change without volume change. Bulk modulus applies to fluids and solids experiencing hydrostatic pressure; shear modulus applies to tangential forces. NEET exam questions from 2014-2025 frequently compare these elastic constants to test conceptual clarity on material rigidity under different stress conditions.
5. How should I approach Poisson's ratio problems in mechanical properties of solids questions for NEET?
Ans. Poisson's ratio (σ) is the negative ratio of lateral strain to longitudinal strain when material is stretched. When a rod elongates, it simultaneously contracts laterally; this ratio quantifies that relationship. NEET previous year questions test Poisson's ratio through problems involving wire deformation and relationships between elastic constants. Typical values range from 0 to 0.5 depending on material type and atomic structure.
Explore Courses for NEET exam
Get EduRev Notes directly in your Google search
Related Searches
NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids, Important questions, Summary, Exam, Semester Notes, Free, shortcuts and tricks, ppt, MCQs, past year papers, Previous Year Questions with Solutions, Viva Questions, NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids, video lectures, study material, Extra Questions, practice quizzes, mock tests for examination, Sample Paper, pdf , Objective type Questions, NEET Previous Year Questions (2014-2025): Mechanical Properties of Solids;