Blog Log in Get Started Free

Mechanical Properties of Solids MCQs for NEET — Physics Questions with Answers

Practice free Mechanical Properties of Solids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

When 100 N tensile force is applied to a rod of $ 10^{-6} m^2 $ cross-sectional area, its length increases by 1% so young's modulus of material is..........

A. $ 10 ^ {12} Pa $ B. $ 10 ^ {11} Pa $ C. $ 10 ^ {10} Pa $ D. $ 10 ^ {2} Pa $

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

A rubber ball when taken to the bottom of a 100 mdeep take decrease in volume by 1% Hence, the bulk modulus of rubber is............$ ( g = 10 m/s^2 ) $

A. $ 10 ^ 6 Pa $ B. $ 10 ^ 8 Pa $ C. $ 10 ^ 7 Pa $ D. $ 10 ^ 9 Pa $

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Bulk modulus (K) is given by the formula: $$ K = rac{Pressure}{rac{ ext{Î”}V}{V}} $$ Pressure (P) at a depth h is given by: $$ P = ho g h $$ Assuming the density of water ( ext{Ï}) to be $1000 ext{ kg/m}^3$: $$ P = 1000 ext{ kg/m}^3 imes 10 ext{ m/s}^2 imes 100 ext{ m} = 10^6 ext{ Pa} $$ The fractional change in volume is 1%, or 0.01. Hence, $$ K = rac{10^6 ext{ Pa}}{0.01} = 10^8 ext{ Pa} $$ Therefore, the correct answer is $10^8 ext{ Pa}$, which corresponds to o2.

● ● ● ● ●

Young's modulus of a rigid body is

A. 0 B. 1 C. $ \infty $ D. 0.5

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Young's modulus (Y) is a measure of the stiffness of a material. For a perfectly rigid body, there is no deformation regardless of the applied force, meaning the strain is zero. Since Young's modulus is calculated as: $$ Y = rac{Stress}{Strain} $$ If strain is zero, the modulus tends to infinity. Therefore, the correct answer is $$ ext{âˆž} $$ which corresponds to o3.

● ● ● ● ●

Pressure on an object increases from $ 1.01 \times 10^ 5 Pa to 1.165 \times 10^ 5 Pa $. He volume decrease by 10% at constant temperature. Bulk modulus of material is........

A. $ 1.55 \times 10 ^ 5 Pa $ B. $ 51.2 \times 10 ^ 5 Pa $ C. $ 102.4 \times 10 ^ 5 Pa $ D. $ 204.8 \times 10 ^ 5 Pa $

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Cross-sectional area if wire of length Lis A. Young's modulus of material is Y. If this wire acts as a spring what is the value of force constant ?

A. YA/L B. YA/2L C. 2YA/L D. YL/A

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$ Y = { F \over A} { L \over \triangle L } $

● ● ● ● ●

A long spring is stretched by 2 cm, its potential energy is U. If the spring is streched by 10 cm, find the potential energy stored in it. [This question is only for Dropper and XII batch]

10 U

15 U

20 U

25 U

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Elastic potential energy of a spring $U = \frac{1}{2} k x^{2} ∴ U \propto x^{2}$

So $\frac{U_{2}}{U_{1}} = {(\frac{x_{2}}{x_{1}})}^{2} \Rightarrow \frac{U_{2}}{U} = {(\frac{10 c m}{2 c m})}^{2} \Rightarrow U_{2} = 25 U$

● ● ● ● ●

A spring of spring constant $5 \times 10^{3} N / m$ is stretched initially by 5 cm from the unstretched position. Find the work required to stretch it further by another 5 cm is [This question is only for Dropper and XII batch]

15 J

18.75 N .m

20 J

22.75 N .m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Work done to stretch the spring from $x_{1} t o x_{2}$

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} 5 \times 10^{3} [{(10 \times 10^{- 2})}^{2} - {(5 \times 10^{- 2})}^{2}] = \frac{1}{2} \times 5 \times 10^{3} \times 75 \times 10^{- 4} = 18.75 N . m .$

● ● ● ● ●

The breaking stress of a wire depends upon

material of the wire

length of the wire

radius of the wire

shape of the cross section

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Breaking stress depends only on nature of material

● ● ● ● ●

The elastic energy stored in a wire of Young's Modulus Y is -

$Y \times \frac{{strain}^{2}}{volume}$

$stress \times strain \times volume$

$\frac{{strain}^{2} \times volume}{2 Y}$

$\frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$U = \frac{1}{2} F ∆ l = \frac{1}{2} σ A ∆ l = \frac{1}{2} σ \frac{∆ l}{l} A l = \frac{1}{2} σ ε V = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$

● ● ● ● ●

If Young modulus (Y) equal to bulk modulus (B). Then the Poisson ratio is :

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Y = 3 B (1 - 2 σ)$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Mechanical Properties of Solids question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.