Mechanical Properties of Solids is an important conceptual and scoring chapter in Class 11 Physics that explains how solids respond to applied forces. This chapter covers key topics such as elastic behaviour, stress and strain, Hooke’s law, stress–strain curve, Young’s modulus, bulk modulus, shear modulus, Poisson’s ratio, and elastic energy, which are frequently asked in school exams and competitive exams like JEE and NEET.
NCERT Solutions for Class 11 Physics Chapter 8 – Mechanical Properties of Solids are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear formulas, graphs, and solved numericals, helping students build strong conceptual clarity and score well in Class 11 examinations.
NCERT Solutions for Class 11 Physics Chapter 8 – Mechanical Properties of Solids
Q.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Q.
Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume =100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Q.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
Q.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.
Q.
Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
Q.
How much should the pressure on a litre of water be changed to compress it by 0.10%?
Q.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×106 Pa.
Q.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Q.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 ×103 kg m–3?
Q.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev s-1 at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Q.
Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Q.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Q.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 Nm–2, what is the maximum load the cable can support?
Q.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Q.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Q.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Q.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Q.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Q.
The stress-strain graphs for materials A and B are shown in Fig. 9.12.
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Q.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 ×107 Pa? Assume that each rivet is to carry one quarter of the load.
NCERT Solutions for Class 11 Physics Chapter 8 – Mechanical Properties of Solids
Q1. Define stress and strain. Mention their SI units.
Answer:
Stress is the internal restoring force per unit area developed in a body when deformed (SI unit: pascal, Pa).
Strain is the fractional change in dimension (ratio), so it is dimensionless (no unit).
Q2. What is Hooke’s law? State the condition under which it holds.
Answer:
Hooke’s law states that within the elastic limit, stress is directly proportional to strain.
It holds only for small deformations where the stress–strain graph remains linear.
Q3. Differentiate between elastic behaviour and plastic behaviour.
Answer:
In elastic behaviour, the body regains its original shape after removing the load.
In plastic behaviour, permanent deformation remains even after unloading.
Q4. Explain Young’s modulus and write its formula.
Answer:
Young’s modulus (Y) measures the stiffness of a material under longitudinal stress.
Y = (F/A)/(ΔL/L).
Q5. What is bulk modulus? What does a higher bulk modulus indicate?
Answer:
Bulk modulus (K) relates pressure change to volume strain: K = −ΔP/(ΔV/V).
A higher K means the material is less compressible.
Q6. Define shear modulus (modulus of rigidity) and give its expression.
Answer:
Shear modulus (G) relates shear stress to shear strain: G = (shear stress)/(shear strain).
It indicates resistance to change in shape at constant volume.
Q7. What is Poisson’s ratio? Give its typical range for most materials.
Answer:
Poisson’s ratio (ν) is the ratio (magnitude) of lateral strain to longitudinal strain.
For most common solids, ν typically lies between 0 and 0.5.
Q8. Why are the elastic moduli different for the same material?
Answer:
They represent resistance to different deformations: length change (Y), volume change (K), and shape change (G).
A material can resist one type more than another.
Q9. Explain the stress–strain curve for a ductile material (key points).
Answer:
Key regions: linear elastic region (Hooke’s law), yield point (plastic deformation begins),
strain hardening up to ultimate strength, then necking and fracture.
Q10. What is the elastic limit? How is it related to yield strength?
Answer:
Elastic limit is the maximum stress without permanent deformation.
Yield strength is the stress where noticeable plastic deformation starts; it is often close to the elastic limit.
Q11. Define ultimate tensile strength and breaking stress.
Answer:
Ultimate tensile strength (UTS) is the maximum stress a material can sustain.
Breaking stress is the stress at fracture (often less than UTS due to necking).
Q12. What is the factor of safety? Why is it used in design?
Answer:
Factor of safety = (maximum/ultimate stress)/(working stress).
It provides a safety margin against uncertainties like extra loads, defects, and fatigue.
Q13. Define elastic energy stored in a stretched wire. What is energy density?
Answer:
Within elastic limit, elastic energy stored: U = 1/2 × (stress) × (strain) × (volume).
Energy density = U/volume = 1/2 × stress × strain.
Q14. Why are bridges and aircraft structures designed keeping fatigue in mind?
Answer:
Cyclic loading can cause failure at stresses lower than UTS due to fatigue crack growth.
So designs account for fatigue limits to prevent long-term sudden failure.
Q15. Why is steel preferred over rubber for making springs?
Answer:
Spring materials need high elastic limit and high Young’s modulus for reliable energy storage and recovery.
Steel satisfies these better than rubber, which shows large hysteresis losses.
FAQs: Class 11 Physics Chapter 8 – Mechanical Properties of Solids
Q1. Is Mechanical Properties of Solids important for exams?
Yes, it is a high-scoring mechanics chapter for Class 11.
Q2. Which topics are most important in this chapter?
Stress–strain curve, elastic moduli, Hooke’s law, and elastic energy.
Q3. Are numericals asked from this chapter?
Yes, Young’s modulus and strain-based numericals are common.
Q4. Are graphs important here?
Yes, the stress–strain graph is frequently asked in exams.
Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals.