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Ever wondered why bridges stay strong under heavy loads or how a rubber band stretches and comes back to its original shape? The chapter Mechanical Properties of Solids in Class 11 Physics explains all of this in a simple and interesting way. It teaches how solid materials behave when forces are applied to them like stretching, compressing or bending.In this article you we get the detailed NCERT solution for class 11 physics chapter 8, These solutions are prepared by our expert faculty as per the latest CBSE syllabus.
This chapter mainly focuses on elasticity and deformation, which are very useful in real life and very important for exams like NEET and JEE Main. If the topics feel hard, these easy to understand NCERT solutions from Careers360 will help you to learn the concept better and solve questions easily. These solutions explain everything step by step so you can learn concept better and solve problems with confidence.
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Get detailed NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids. Download the free PDF below and understand every concept with step-by-step answers.
Answer:
Let the Young's Modulus of steel and copper be Y S and Y C respectively.
Length of the steel wire l S = 4.7 m
Length of the copper wire l C = 4.7 m
The cross-sectional area of the steel wire A S =
The cross-sectional area of the Copper wire A C =
Let the load and the change in the length be F and
Since F and
The ratio of Young’s modulus of steel to that of copper is 1.79.
8.2 (a) Figure 8.9 shows the strain-stress curve for a given material. What are (a)
Fig. 8.9
Answer:
Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.
For the given material
8.2(b)What are approximate yield strength for this material?
Answer:
The Yield Strength is approximately
Fig 8.10
Answer:
As we can see in the given Stress-Strain graphs that the slope is more in the graph corresponding to material A. We conclude that A has a greater Young's Modulus.
8.3(b) Which of the two is the stronger material?
Answer: The material which fractures at higher stress is said to be stronger. As we can see in the given Stress-Strain graphs the stress at which the material fractures is higher in A than that in B we conclude that A is the stronger material.
8.4(a) Read the following two statements below carefully and state, with reasons, if it is true or false.
The Young’s modulus of rubber is greater than that of steel;
Answer:
False: Young's Modulus is defined as the ratio of the stress applied on a material and the corresponding strain that occurs. As for the same amount of pressure applied on a piece of rubber and steel, the elongation will be much lesser in case of steel than that in the case of rubber and therefore the Young's Modulus of rubber is lesser than that of steel.
Answer:
True: As the force acts Normal to the parallel planes in which helical parts of the wire lie, the actual length of the wire would not change but it's shape would. Therefore the amount of elongation of the coil taking place for corresponding stress depends upon the Shear Modulus of elasticity.
Fig 8.11
Answer:
Tension in the steel wire is F 1
Length of steel wire l 1 = 1.5 m
The diameter of the steel wire, d = 0.25 cm
Area od the steel wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y 1 =
Tension in the Brass wire is F 2
Length of Brass wire l 2 = 1.5 m
Area of the brass wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y 2 =
Answer:
Edge of the aluminium cube, l = 10 cm = 0.1 m
Area of the face of the Aluminium cube, A = l 2 = 0.01 m 2
Tangential Force is F
Tangential Stress is F/A
Shear modulus of aluminium
Let the Vertical deflection be
Answer:
Inner radii of each column, r 1 = 30 cm = 0.3 m
Outer radii of each column, r 2 = 60 cm = 0.6 m
Mass of the structure, m = 50000 kg
Stress on each column is P
Youngs Modulus of steel is
Answer:
Length of the copper piece, l = 19.1 mm
The breadth of the copper piece, b = 15.2 mm
Force acting, F = 44500 N
Modulus of Elasticity of copper,
Answer:
Let the maximum Load the Cable Can support be T
Maximum Stress Allowed, P = 10 8 N m -2
Radius of Cable, r = 1.5 cm
Answer:
Each wire must support the same load and are of the same length and therefore should undergo the same extension. This, in turn, means they should undergo the same strain.
Answer:
Mass of the body = 14.5 kg
Angular velocity,
The radius of the circle, r = 1.0 m
Tension in the wire when the body is at the lowest point is T
Cross-Sectional Area of wire, A = 0.065 cm 2
Young's Modulus of steel,
Answer:
Pressure Increase, P = 100.0 atm
Initial Volume = 100.0 l
Final volume = 100.5 l
Change in Volume = 0.5 l
Let the Bulk Modulus of water be B
The bulk modulus of air is
The Ratio of the Bulk Modulus of water to that of air is
This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water.
Answer:
Water at the surface is under 1 atm pressure.
At the depth, the pressure is 80 atm.
Change in pressure is
Bulk Modulus of water is
The negative sign signifies that for the same given mass the Volume has decreased The density of water at the surface
Let the density at the given depth be
Let a certain mass occupy V volume at the surface
Dividing the numerator and denominator of RHS by V we get
The density of water at a depth where pressure is 80.0 atm is
Answer:
Bulk's Modulus of Glass is
Pressure is P = 10 atm.
The fractional change in Volume would be given as
The fractional change in Volume is
Answer:
The bulk modulus of copper is
Edge of copper cube is s = 10 cm = 0.1 m
Volume Of copper cube is V = s 3
V = (0.1) 3
V = 0.001 m 3
Hydraulic Pressure applies is
From the definition of bulk modulus
The volumetric strain is
Volume contraction will be
The volume contraction has such a small value even under high pressure because of the extremely large value of the bulk modulus of copper.
8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer:
Change in volume is
Bulk modulus of water is
A pressure of
Note: The answer is independent of the volume of water taken into consideration. It only depends on the percentage change.
1. A rigid bar of mass M is supported symmetrically by three wires, each of length L. Those at each end are of copper, and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to?
Solution:
Hence,
So,
2. What will be the maximum load a wire can withstand without breaking when its length is reduced to half of its original length?
Solution: Stress = force/area
The breaking stress is not dependent on the length, and hence, if the cross-sectional area is changed, it will not affect the breaking force.
3.The temperature of a wire is doubled. The Young's modulus of elasticity
a) will also double
b) will become four times
c) will remain same
d) will decrease
Solution: As we learn,
So as the Young’s modulus increases, elasticity decreases.
4. A spring is stretched by applying a load to its free end. What is the strain produced in the spring?
Solution: The strain produced in a material can be shearing or longitudinal, depending on the type of force applied. In the case of a spring stretched by a load, the shape and length of the spring both change. When a longitudinal strain is produced, the length of the spring increases or decreases in the direction of the applied force.
5. Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both rods are fixed rigidly at one end of the roof. A mass M is attached to each of the free ends at the centre of the rods. Then:
a) both the rods will elongate but there shall be no perceptible change in shape
b) the steel rod will elongate and change shape but the rubber rod will only elongate
c) the steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse
d) the steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre
Solution: The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
Y steel
Hence,
To solve the questions from the chapter Mechanical Properties of Solids make sure that you understand the basic concepts like stress, strain, Young’s modulus, bulk modulus, shear modulus and Hooke’s law. Read each question carefully and find what is given and what to calculate. Use the right formula based on the given information. Also, make sure that units must be same throughout the calculation. Try to draw diagrams(FBD) to simplify the question. Practice numerical problems regularly to improve your concepts and calculation speed.
Concept Name | JEE | NCERT |
Elasticity | ✅ | ✅ |
Stress And Strain | ✅ | ✅ |
Stress Strain Relationship | ✅ | ✅ |
Hooke’s Law | ✅ | ✅ |
Work Done In Stretching A Wire | ✅ | ✅ |
Relation Between Volumetric Strain, Lateral Strain And Poisson’s Ratio | ✅ | ❌ |
The topics covered in the NCERT book Class 11 physics chapter 9 are
One question may be asked from mechanical properties of solids for JEE Main exam. To practice more questions refer to previous year JEE main papers, NCERT book and NCERT exemplar problems.
Use the Mechanical Properties of Solids Class 11 NCERT PDF to revise topics like stress, strain, elasticity, and Hooke’s Law. It's a handy tool for practicing questions before tests and clearing your basics.
You can easily download the Mechanical Properties of Solids NCERT Solutions PDF from the above-given download link. These solutions help you understand key concepts and solve textbook questions with step-by-step answers.
Stress is the force applied per unit area on a material, causing it to deform.
Exam Date:22 July,2025 - 29 July,2025
Exam Date:22 July,2025 - 28 July,2025
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