Chapter 8 Mechanical Properties of Solids

NCERT Solutions Class 11 Physics Chapter 8 – Mechanical Properties of Solids

NCERT Solutions for Class 11 Physics Chapter 8 (Mechanical Properties of Solids) transition from the study of rigid bodies to Deformable Bodies. In reality, nobody is perfectly rigid; when an external force is applied, objects undergo changes in shape or size. This chapter explores the molecular basis of Elasticity—the property by which a body returns to its original dimensions after the deforming force is removed.

The NCERT Solutions for Class 11 Physics Chapter 8 provided by ALLEN are formulated by expert faculty to explain the relationships between stress and strain and elastic constants in a lucid way. The stepwise solutions are designed to meet the problem solving requirements of CBSE, JEE and NEET.

A good understanding of the behaviour of materials under stress is fundamental for civil and mechanical engineers. These solutions offer a simple way to figure out how much a wire stretches or a pillar compresses, which is necessary information for designing safe buildings, bridges, and machinery.

1.0Class 11 Physics Chapter 8 : Key Concepts

This chapter examines how solids respond to external forces and the limits of their structural integrity. Key lessons include:

Elastic Behavior of Solids: Understanding the intermolecular forces that act like "springs" between atoms.
Stress and Strain:

Stress: The internal restoring force per unit area ( $σ = F / A$ ). Types include Tensile, Compressive, and Tangential (Shear) stress.
Strain: The fractional change in dimensions. Types include Longitudinal $(Δ L / L),$ Shearing $(θ)$ , and Volumetric $(Δ V / V)$ .
Hooke’s Law: For small deformations, stress is directly proportional to strain ( $St ress = E \times St r ain$ ).
Stress-Strain Curve: Analyzing the behavior of a material from the elastic limit to the fracture point.

Elastic Moduli:

Young’s Modulus (Y): Measure of resistance to longitudinal change $(Y = \frac{M gL}{π r ^{2} Δ L}) .$
Bulk Modulus (B): Measure of resistance to volume change $(B = - P / (Δ V / V)) .$
Shear Modulus (G): Also known as Modulus of Rigidity, measuring resistance to change in shape.
Applications of Elasticity: Elasticity is used in understanding why I-shaped beams are used in construction and the maximum height of a mountain on Earth.

Elastic Potential Energy: The energy stored in a stretched wire $(U = \frac{1}{2} \times St ress \times St r ain \times V o l u m e)$ .

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 8

SOLVED EXAMPLES

A structural steel rod has a radius of 10 mm and a length of 1.0m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation and (c) strain on the rod. Young's modulus of structural steel is $2.0 \times 1 0^{11} Nm^{- 2}$ . Sol. We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by

$Stress = \frac{F}{A} = \frac{F}{π r ^{2}} = \frac{100 \times 1 0 ^{3} N}{3.14 \times ( 1 0 ^{- 2} m ) ^{2}} = 3.18 \times 1 0^{8} Nm^{- 2}$

The elongation,

$Δ L = \frac{( F / A ) L}{Y} = \frac{( 3.18 \times 1 0 ^{8} Nm ^{- 2} ) ( 1 m )}{2 \times 1 0 ^{11} Nm ^{- 2}} = 1.59 \times 1 0^{- 3} m = 1.59 mm$

The strain is given by

$Strain = \frac{Δ L}{L} = (1.59 \times 1 0^{- 3} m) / (1 m) = 1.59 \times 1 0^{- 3} = 0.16%$

A copper wire of length 2.2 m and a steel wire of length 1.6 m , both of diameter 3.0 mm , are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied. Sol. The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. From Eq. we have stress = strain × Young's modulus. Therefore $\frac{W}{A} = Y_{c} \times \frac{Δ L _{c}}{L _{c}}$

$= Y_{s} \times \frac{Δ L _{S}}{L _{S}}$

where the subscripts c and s refer to copper and stainless steel respectively. Or,

$\frac{Δ L _{c}}{Δ L _{s}} = \frac{Y _{s}}{Y _{c}} \times \frac{L _{c}}{L _{s}}$

Given $L_{c} = 2.2 m, L_{s} = 1.6 m$

$Y_{c} = 1.1 \times 1 0^{11} N . m^{- 2} and Y_{s} = 2.0 \times 1 0^{11} N . m^{- 2} \frac{Δ L _{c}}{Δ L _{S}} = \frac{2.0 \times 1 0 ^{11}}{1.1 \times 1 0 ^{11}} \times (\frac{2.2}{1.6}) = 2.5$

The total elongation is given to be

$Δ L_{c} + Δ L_{s} = 7.0 \times 1 0^{- 4} m$

Solving the above equations,

$Δ L_{c} = 5.0 \times 1 0^{- 4} m and Δ L_{s} = 2.0 \times 1 0^{- 4} m$

Therefore

$W = \frac{( A \times Y _{c} \times Δ L _{c} )}{L _{c}} = \frac{π ( 1.5 \times 1 0 ^{- 3} ) ^{2} \times [ 5.0 \times 1 0 ^{- 4} \times 1.1 \times 1 0 ^{11} )}{2.2} = 1.8 \times 1 0^{2} N$

In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in figure). The combined mass of all the persons performing the act and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Sol. Total mass of all the performers, tables, plaques etc. $= 280 kg$ Mass of the performer $= 60 kg$ Mass supported by the legs of the performer at the bottom of the pyramid

$= 280 - 60 = 220 kg$

Weight of this supported mass

$= 220 kg wt. = 220 \times 9.8 N = 2156 N$

Weight supported by each thighbone of the

$performer = \frac{1}{2} (2156) N = 1078 N$

Young's modulus for bone is given by

$Y = 9.4 \times 1 0^{9} N m^{- 2}$

Length of each thighbone $L = 0.5 m$ the radius of thighbone $= 2.0 cm$ Thus the cross-sectional area of the thighbone

$A = π \times (2 \times 1 0^{- 2})^{2} m^{2} = 1.26 \times 1 0^{- 3} m^{2}$

compression in each thighbone ( $Δ L$ ) can be computed as

$Δ L = [\frac{( F \times L )}{( Y \times A )}] = \frac{( 1078 \times 0.5 )}{( 9.4 \times 1 0 ^{9} \times 1.26 \times 1 0 ^{- 3} )} = 4.55 \times 1 0^{- 5} m or 4.55 \times 1 0^{- 3} cm$

This is a very small change! The fractional decrease in the thighbone is $Δ L / L = 0.000091$ or 0.0091%.

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of $9.0 \times 1 0^{4} N$ . The lower edge is riveted to the floor. How much will the upper edge be displaced? Sol. The lead slab is fixed and the force is applied parallel to the narrow face as shown in figure. The area of the face parallel to which this force is applied is

$A = 50 cm \times 10 cm = 0.5 m \times 0.1 m = 0.05 m^{2}$

Therefore, the stress applied is $= \frac{F}{A}$

$= (\frac{9.0 \times 1 0 ^{4} N}{0.05 m ^{2}}) = 1.80 \times 1 0^{6} N . m^{- 2}$

We know that shearing strain

$= \frac{Δ x}{L} = \frac{Stress}{G}$

Therefore the displacement

$Δ x = \frac{Stress \times L}{G} = \frac{( 1.8 \times 1 0 ^{6} Nm ^{- 2} \times 0.5 m )}{( 5.6 \times 1 0 ^{9} Nm ^{- 2} )} = 1.6 \times 1 0^{- 4} m = 0.16 mm$

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, $Δ V / V$ , of water at the bottom of the ocean, given that the bulk modulus of water is $2.2 \times 1 0^{9} Nm^{- 2}$ . (Take $g = 10 ms^{- 2}$ ) Sol. The pressure exerted by a 3000 m column of water on the bottom layer $p = h ρ g = 3000 m \times 1000 kg m^{- 3} \times 10 ms^{- 2}$ $= 3 \times 1 0^{7} kg m^{- 1} s^{- 2} = 3 \times 1 0^{7} Nm^{- 2}$ Fractional compression $\frac{Δ V}{V}$ , is

$\frac{Δ V}{V} = stress / B = \frac{( 3 \times 1 0 ^{7} Nm ^{- 2} )}{( 2.2 \times 1 0 ^{9} Nm ^{- 2} )} = 1.36 \times 1 0^{- 2} or 1.36%$

EXERCISE QUESTIONS WITH SOLUTIONS

A steel wire of length 4.7 m and crosssectional are $3.0 \times 1 0^{- 5} m^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 1 0^{- 5} m^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper? Sol. Length of the steel wire, $L_{1} = 4.7 m$ Area of cross-section of the steel wire, $A_{1} = 3.0 \times 1 0^{- 5} m^{2}$ Length of the copper wire, $L_{2} = 3.5 m$ Area of cross-section of the coper wire, $A_{2} = 4.0 \times 1 0^{- 5} m^{2}$ Change in length $= Δ L_{1} = Δ L_{2} = Δ L$ Force applied in both the cases = F Young's modulus of the steel wire :

$Y_{1} = (\frac{F _{1}}{A _{1}}) (\frac{L _{1}}{Δ L _{1}}) = (\frac{F}{3 \times 1 0 ^{- 5}}) (\frac{4.7}{Δ L})$

Young's modulus of the copper wire :

$Y_{2} = (\frac{F _{2}}{A _{2}}) (\frac{L _{2}}{Δ L _{2}}) = (\frac{F}{4 \times 1 0 ^{- 5}}) (\frac{3.5}{Δ L})$

Dividing (i) by (ii), we get :

$\frac{Y _{1}}{Y _{2}} = \frac{( 4.7 \times 4 \times 1 0 ^{- 5} )}{( 3 \times 1 0 ^{- 5} \times 3.5 )} = 1.79$

The ratio of Young's modulus of steel to that of copper is 1.79 : 1.

2. Figure shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?

Sol. (a) It is clear from the given graph that for stress $150 \times 1 0^{6} N / m^{2}$ , strain is 0.002 . ∴ Young's modulus,

$Y = \frac{Stress}{Strain} = \frac{150 \times 1 0 ^{6}}{0.002} = 7.5 \times 1 0^{10} Nm^{- 2}$

Hence, Young's modulus for the given material is $7.5 \times 1 0^{10} N / m^{2}$ . (b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is clear from the given graph that the approximate yield strength of this material is $300 \times 1 0^{6} N / m^{2}$ or $3 \times 1 0^{8} N / m^{2}$ .

The stress-strain graphs for materials A and B are shown in figure.

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?

Sol. (a) From the two graphs, we note that for a given strain, stress of A is more than that of B. Hence, Young's modulus (=stress/strain) is greater for A than that of B. (b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

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. Sol. (a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain. (b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. 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.

Sol. Diameter of the wires, $d = 0.25 m$ Hence, the radius of the wires,

$r = \frac{d}{2} = 0.125 cm$

Length of the steel wire, $L_{1} = 1.5 m$ Length of the brass wire, $L_{2} = 1.0 m$ Total force exerted on the steel wire :

$F_{1} = (4 + 6) g = 10 \times 9.8 = 98 N$

Young's modulus for steel :

$Y_{1} = \frac{F _{1}}{A _{1}} \times \frac{L _{1}}{Δ L _{1}}$

Where,

$Δ L_{1} = Change in the length of the steel wire A_{1} = Area of cross-section of the steel wire = π r_{1}^{2}$

Young's modulus of steel,

$Y_{1} = 2.0 \times 1 0^{11} Pa ∴ Δ L_{1} = \frac{F _{1} \times L _{1}}{A _{1} \times Y _{1}} = \frac{98 \times 1.5}{[ π ( 0.125 \times 1 0 ^{- 2} ) ^{2} \times 2 \times 1 0 ^{11} ]} = 1.49 \times 1 0^{- 4} m$

Total force on the brass wire :

$F_{2} = 6 \times 9.8 = 58.8 N$

Young's modulus for brass :

$Y_{2} = \frac{F _{2}}{A _{2}} \times \frac{L _{2}}{Δ L _{2}}$

Where,

$Δ L_{2} = Change in the length of the brass wire A_{2} = Area of cross-section of the brass wire = π r_{2}^{2}$

$∴ Δ L_{2} = \frac{F _{2} \times L _{2}}{A _{2} \times Y _{2}} = \frac{58.8 \times 1}{π ( 0.125 \times 1 0 ^{- 2} ) ^{2} \times ( 0.91 \times 1 0 ^{11} )} = 1.3 \times 1 0^{- 4} m$

Elongation of steel wire $= 1.49 \times 1 0^{- 4} m$ Elongation of the brass wire $= 1.3 \times 1 0^{- 4} m$

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? Sol. Edge of the aluminium cube,

$L = 10 cm = 0.1 m$

The mass attached to the cube, $m = 100 kg$ Shear modulus ( $η$ ) of

$aluminium = 25 GPa = 25 \times 1 0^{9} Pa$

Shear modulus, $η =$ Shear stress/Shear strain

$= \frac{F}{A} \times \frac{L}{Δ L}$

Where,

$F = Applied force = m g = 100 \times 9.8 = 980 N$

A = Area of one of the faces of the cube

$= 0.1 \times 0.1 = 0.01 m^{2}$

$Δ L =$ Vertical deflection of the cube

$∴ Δ L = \frac{FL}{A η} = \frac{980 \times 0.1}{1 0 ^{- 2} \times 25 \times 1 0 ^{9}} = 3.92 \times 1 0^{- 7} m$

The vertical deflection of this face of the cube is $3.92 \times 1 0^{- 7} m$

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. Sol. Mass of the big structure,

$M = 50, 000 kg$

Inner radius of the column,

$r = 30 cm = 0.3 m$

Outer radius of the column,

$R = 60 cm = 0.6 m$

Young's modulus of steel,

$Y = 2 \times 1 0^{11} a$

Total force exerted,

$F = Mg = 50000 \times 9.8 N$

$Stress = Force exerted on a single column$

$= \frac{50000 \times 9.8}{4} = 122500 N$

Young's modulus, $Y = \frac{Stress}{Strain}$

$Strain = \frac{F}{AY}$

Where, Area, $A = π (R^{2} - r^{2}) = π [(0.6)^{2} - (0.3)^{2}]$ Strain

$= \frac{122500}{[ π { ( 0.6 ) ^{2} - ( 0.3 ) ^{2} } \times 2 \times 1 0 ^{11} ]} = 7.22 \times 1 0^{- 7}$

Hence, the compressional strain of each column is $7.22 \times 1 0^{- 7}$ .

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? Sol. Length of the piece of copper,

$l = 19.1 mm = 19.1 \times 1 0^{- 3} m$

Breadth of the piece of copper,

$b = 15.2 mm = 15.2 \times 1 0^{- 3} m$

Area of the copper piece :

$A = l \times b = 19.1 \times 1 0^{- 3} \times 15.2 \times 1 0^{- 3} = 2.9 \times 1 0^{- 4} m^{2}$

Tension force applied on the piece of copper,

$F = 44500 N$

Modulus of elasticity of copper,

$η = 42 \times 1 0^{9} N / m^{2}$

Modulus of elasticity, $η = \frac{Stress}{Strain}$

$= \frac{\frac{F}{A}}{Strain} ∴ Strain = \frac{F}{A η} = \frac{44500}{( 2.9 \times 1 0 ^{- 4} \times 42 \times 1 0 ^{9} )} = 3.65 \times 1 0^{- 3} .$

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 $1 0^{8} Nm^{- 2}$ , What is the maximum load the cable can support? Sol. Radius of the steel cable,

$r = 1.5 cm = 0.015 m Maximum allowable stress = 1 0^{8} Nm^{- 2}$

$Maximum stress = \frac{Maximum force}{Area of cross - section}$

$∴ Maximum force =$

Maximum stress × Area of cross section

$= 1 0^{8} \times π (0.015)^{2} = 7.065 \times 1 0^{4} N$

Hence, the cable can support the maximum load of $7.065 \times 1 0^{4} N$

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of coper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension. Sol. The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same. The relation for Young's modulus is given as :

$Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{Strain} = \frac{\frac{4 F}{π d ^{2}}}{Strain}$

Where,

$F = Tension force A = Area of cross-section d = Diameter of the wire$

It can be inferred from equation (i) that

$Y \propto (1/ d^{2})$

Young's modulus for iron,

$Y_{1} = 190 \times 1 0^{9} Pa$

Diameter of the iron wire $= d_{1}$ Young's modulus for copper, $Y_{2} = 120 \times 1 0^{9} Pa$ Diameter of the coper wire $= d_{2}$ Therefore, the ratio of their diameters is given as :

$\frac{d _{1}}{d _{2}} = \frac{Y _{1}}{Y _{2}} = \frac{190 \times 1 0 ^{9}}{120 \times 1 0 ^{9}} = \frac{19}{12} = 1.25$

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 at the bottom of the circle. The crosssectional area of the wire is $0.065 cm^{2}$ . Calculate the elongation of the wire when the mass is at the lowest point of its path. Sol. Mass, $m = 14.5 kg$ Length of the steel wire, $l = 1.0 m$ Angular velocity,

$ω = 2 rev / s = 2 \times 2 π rad / s = 12.56 rad / s$

Cross-sectional area of the wire,

$a = 0.065 cm^{2} = 0.065 \times 1 0^{- 4} m^{2}$

Let $Δ l$ be the elongation of the wire when the mass is the lowest point of its path. When the mass is placed at the position of the vertical circle, the total force on the mass is:

$F = mg + m l ω^{2} = 14.5 \times 9.8 + 14.5 \times 1 \times (12.56)^{2} = 2429.53 N$

Young's modulus $= \frac{Stress}{Strain}$

$Y = \frac{F}{A} \times \frac{l}{Δ l} ∴ Δ l = \frac{F l}{AY}$

Young's modulus for steel $= 2 \times 1 0^{11} Pa$

$Δ l = \frac{2429.53}{( 0.065 \times 1 0 ^{- 4} \times 2 \times 1 0 ^{11} )} = 1.87 \times 1 0^{- 3} m$

Hence, the elongation of the wire is

$1.87 \times 1 0^{- 3} m .$

Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure increase $= 100.0 atm$ $(1 atm = 1.013 \times 1 0^{5} Pa)$ . 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. Sol. Initial volume, $V_{1} = 100.0$ litre $= 100.0 \times 1 0^{- 3} m^{3}$ Final volume, $V_{2} = 100.5$ litre $= 100.5 \times 1 0^{- 3} m^{3}$ Increase in volume, $Δ V = V_{2} - V_{1} = 0.5 \times 1 0^{- 3} m^{3}$ Increase in pressure,

$Δ p = 100.0 atm = 100 \times 1.013 \times 1 0^{5} Pa$

Bulk modulus $= \frac{Δ p}{\frac{Δ V}{V _{1}}} = \frac{Δ p \times V _{1}}{Δ V}$ $= \frac{100 \times 1.013 \times 1 0 ^{5} \times 100 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}}$ $= 2.026 \times 1 0^{9} Pa$ Bulk modulus of air $= 1 \times 1 0^{5} Pa$

$∴ \frac{Bulk modulus of water}{Bulk modulus of air} = \frac{2.026 \times 1 0 ^{9}}{( 1 \times 1 0 ^{5} )} = 2.026 \times 1 0^{4}$

This ratio is very high because air is more compressible than water.

What is the density of water at a depth where pressure is 80.0 atm , given that its density at the surface is $1.03 \times 1 0^{3} kg m^{- 3}$ ? Sol. Let the given depth be h. Pressure at the given depth,

$p = 80.0 atm = 80 \times 1.01 \times 1 0^{5} Pa$

Density of water at the surface,

$ρ_{1} = 1.03 \times 1 0^{3} kg m^{- 3}$

Let $ρ_{2}$ be the density of water at the depth h. Let $V_{1}$ be the volume of water of mass m at the surface.Let $V_{2}$ be the volume of water of mass at the depth h. Let $Δ V$ be the change in volume.

$Δ V = V_{1} - V_{2} = m [\frac{1}{ρ _{1}} - \frac{1}{ρ _{2}}]$

$∴$ Volumetric strain $= \frac{Δ V}{V _{1}}$

$= m [\frac{1}{ρ _{1}} - \frac{1}{ρ _{2}}] \times \frac{ρ _{1}}{m} = 1 - \frac{ρ _{1}}{ρ _{2}}$

$\frac{Δ V}{V _{1}} = \frac{p}{B}$ Compressibility of water $= (\frac{1}{B})$

$= 45.8 \times 1 0^{- 11} Pa^{- 1} ∴ \frac{Δ V}{V _{1}} = 80 \times 1.013 \times 1 0^{5} \times 45.8 \times 1 0^{- 11} = 3.71 \times 1 0^{- 3}$

For equations (i) and (ii), we get :

$1 - (\frac{ρ _{1}}{ρ _{2}}) = 3.71 \times 1 0^{- 3} ρ_{2} = \frac{1.03 \times 1 0 ^{3}}{[ 1 - ( 3.71 \times 1 0 ^{- 3} ) ]} = 1.034 \times 1 0^{3} kg m^{- 3}$

Therefore, the density of water at the given depth (h) is $1.034 \times 1 0^{3} kg m^{- 3}$ .

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. Sol. Hydraulic pressure exerted on the glass slab, $p = 10 atm = 10 \times 1.013 \times 1 0^{5} Pa$ Bulk modulus of glass, $B = 37 \times 1 0^{9} Nm^{- 2}$ Bulk modulus, $B = \frac{p}{\frac{Δ V}{V}}$ Where, $\frac{Δ V}{V} =$ Fractional change in volume

$∴ \frac{Δ V}{V} = \frac{p}{B} \frac{Δ V}{V} = \frac{10 \times 1.013 \times 1 0 ^{5}}{( 37 \times 1 0 ^{9} )} = 2.73 \times 1 0^{- 5}$

Hence, the fractional change in the volume of the glass slab is $2.73 \times 1 0^{- 5}$ .

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0 \times 1 0^{6} Pa$ . Sol. Length of an edge of the solid copper cube, $l = 10 cm = 0.1 m$ Hydraulic pressure, $p = 7.0 \times 1 0^{6} Pa$ Bulk modulus of coper, $B = 140 \times 1 0^{9} Pa$ Bulk modulus, $B = \frac{p}{\frac{Δ V}{V}} \Rightarrow Δ V = \frac{pV}{B}$ Where, $\frac{Δ V}{V} =$ Volumetric strain $Δ V =$ Change in volume V = Original volume Original volume of the cube, $V = l^{3}$

$∴ Δ V = \frac{p l ^{3}}{B} = \frac{7 \times 1 0 ^{6} \times ( 0.1 ) ^{3}}{( 140 \times 1 0 ^{9} )} = 5 \times 1 0^{- 8} m^{3} = 5 \times 1 0^{- 2} cm^{- 3}$

Therefore, the volume contraction of the solid copper cube is $5 \times 1 0^{- 2} cm^{- 3}$ .

How much should the pressure on a litre of water be changed to compress it by 0.10%? Sol. Volume of water, $V = 1 L$ It is given that water is to be compressed by 0.10%. ∴ Fractional change, $\frac{Δ V}{V} = \frac{0.1}{( 100 \times 1 )} = 1 0^{- 3}$

Bulk modulus, $B = \frac{ρ}{( \frac{Δ V}{V} )}$

$ρ = B \times (\frac{Δ V}{V})$

Bulk modulus of water, $B = 2.2 \times 1 0^{9} Nm^{- 2}$ $ρ = 2.2 \times 1 0^{9} \times 1 0^{- 3} = 2.2 \times 1 0^{6} Nm^{- 2}$ Therefore, the pressure on water should be $2.2 \times 1 0^{6} Nm^{- 2}$ .

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Explore chapter-wise NCERT Solutions for Class 11 Physics with clear explanations for every textbook question. Understand key concepts, follow step-by-step solutions, and strengthen your Physics knowledge for better learning and revision.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 8

Clear Understanding of Stress-Strain Curves: Learn to identify the proportional limit, elastic limit, yield point, and permanent deformation from stress-strain graphs.
Step-by-Step Elasticity Numericals: Solve problems based on Young’s modulus, stretching of wires, compression of materials, stress, and strain with proper formulas, units, and calculations.
Easy Derivation of Elastic Potential Energy: Understand how work done in stretching a wire leads to the expression for elastic potential energy through clear, step-by-step derivations.
Real-Life Applications of Elasticity: Understand the use of elasticity, safety factors, and material strength in applications such as cranes and metallic ropes used for lifting heavy loads.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 8, prepared according to the latest NCERT syllabus.
Simple and Logical Explanation: Know about stress, strain, Hooke’s law, Young’s modulus, bulk modulus, shear modulus and elastic behaviour with easy explanations and stepwise solutions.
Complete Coverage of NCERT Exercises: Understand the concepts of Mechanical Properties of Solids with the help of detailed solutions to NCERT Class 11 Physics Chapter 8 examples and exercise questions for CBSE, JEE and NEET exam preparations.

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NCERT Solutions Class 11 Physics Chapter 8 – Mechanical Properties of Solids

1.0Class 11 Physics Chapter 8 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 8

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 8

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