NCERT Solutions for Class 11 Physics Chapter 9  Mechanical Properties of Solids
NCERT solutions for Class 11 Physics chapter Mechanical Properties of Solids offer simple answers to complex questions. It is imperative for students to prepare well for their exams, and the PDF solutions will help you in this endeavour.
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Chapter Name:  Chapter 9  Mechanical Properties of Solids 
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NCERT for Physics Chapter 9 introduces the Mechanical Properties of Solids. The chapter of Mechanical properties of solids Class 11 explains the prominent characteristics displayed by matters in this stage and goes on to pinpoint the laws governing these situations. It is, therefore, an important chapter for students.
Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.
Mechanical Properties Of Solids Chapter at a Glance  Class 11 NCERT Solutions
Elastic and Plastic behavior of Materials
Whenever a force is applied on a body, then it tends to change the size or shape of the body.
The property of a body by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as Elasticity and the deformation caused is known as Elastic Deformation.
Those substances which do not have a tendency to regain their shape and hence gets permanently deformed are called Plastic and the property is called Plasticity.
Stress
The restoring force per unit crosssectional are set up within the body is called stress.
$Stress\frac{Restoring\; force }{Area\;of\;cross\;section}=\frac{F}{A}$
In SI system, unit of stress is N/m2 or pascal (Pa). In general there are three types of stresses
(a) Longitudinal Stress – Tensile stress (associated with stretching or compressive stress (associated with compression).
(b) Shearing Stress.
(c) Bulk Stress.
Breaking Stress
The minimum stress after which the wire breaks is called breaking stress.
A wire of length l will break due to its own weight, when $l=\frac{Breaking\;stress}{dg}$
where, d = density of material of wire and g = acceleration due to gravity.
Strain
Strain is defined as the ratio of change in dimension of an object to the original dimension.
$Strain=\frac{Change\;in\;configuration\;of\;the\;Object}{Original\;configuration\;of\;the\;Object}$
It is a pure number and has no unit.
Hooke’s Law
This law states that, for small deformations, the stress and strain are proportional to each other.
Thus, Stress\propto Strain
$Stress=k\times\;strain$
The SI unit of modulus of elasticity is Nm2.
A class of solids called elastomers do not obey Hooke’s law.
Moduli of Elasticity
Three elastic moduli viz. Young’s Modulus, shear modulus and bulk modulus are used to describe the elastic behavior of objects as they respond to deforming forces that act on them.
When an object is under tension or compression, the Hooke’s law takes the form:
$ \frac{F}{A}=\frac{Y\Delta L}{L}$
where $\frac{\Delta L}{L}$ is the tensile or compressive strain of the object, F is the magnitude of the applied fore causing the strain. A is the crosssectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object.
A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement $\Delta L$ of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids.
In this kind of deformation the Hooke’s law takes the form :
$\frac{F}{A}=G\times\frac{\Delta L}{L}$ where $\Delta L$ is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus.
When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form $P= B\left (\frac{\Delta V}{V} \right )$. where p is the pressure (hydraulic stress) on the object due to the fluid, $\frac{\Delta V}{V}$ (the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object.
The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes.
Bulk modulus is relevant for solids, liquid and gases. it refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged.
Metal shave larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus require a large force to produce small changes in its length.
Poisson’s Ratio
It is the ratio of transverse or lateral strain to longitudinal strain in the direction of stretching force. It is expressed as
$\sigma =\frac{Lateral\;Contraction\;strain}{Longitudinal\;Contraction\;strain}=\frac{d/D}{l/L}$
where, d = change in diameter & D = original diameter, l change in length & L = original length.
Theoretical limits of Poisson’s ratio are 1 and 0.5, while the practical limits are 0 and 0.5.
Relation between Y,K,G and 𝝈 as given below
Stressstrain Curve
A typical stressstrain curve for a metal is shown in figure
In the region O to A, stress is found proportional to strain. Thus, Hooke’s law is fully obeyed in this region and body regain its original shape.
Point A is known as point of proportional limit.
In the region from A to B, stress and strain are not proportional, but the body still returns to its original shape and size.
The point B is yield point (also called elastic limit) and corresponding stress is yield stress (\sigma y).
If stress increases beyond point B, the strain further increases, but on removing the strain wire does not regain its original length.
Beyond point C, for a small stress, the strain produced is large upto point D. The wire will break at point D called fracture point of wire.
Elastic Potential energy Stored in a Stretched Wire
The work done in stretching a wire against the interatomic forces is stored as the elastic potential energy.
Potential energy of work done per unit is given by
$U=\frac{W}{V}=\frac{1}{2}\times\;Stress\times\;Strain$
$=\frac{1}{2}\times\;Y\;\times\;\left ( Strain \right )^{2}=\frac{1}{2Y}\times\;\left ( Strain \right )^{2}\left [ \vec{} Y=\frac{Stress}{Strain}\right ]$
Potential energy stored in stretching wire by restoring force is expressed as
$W=\frac{1}{2}\times\;F\;\times\;l$
where, F = restoring force and l= elongation produced.
If the force acting on the body is increased from F1 to F2 within the elastic limit, then
$W=\frac{\left ( F_{1}+F_{2} \right )}{2}\times\;extension$
NCERT Solution of Physics Class 11 Chapter 9  Free PDF Download
1. A steel wire of length 4.7 m and crosssectional area \[3.0\times {{10}^{5}}{{m}^{2}}\] stretches by the same amount as a copper wire of length 3.5 m and crosssectional area \[4.0\times {{10}^{5}}{{m}^{2}}\] of under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Ans: In the above question it is given that:
The length of the steel wire is ${{L}_{1}}=4.7m$.
Area of crosssection of the steel wire is \[{{A}_{1}}=3.0\times {{10}^{5}}{{m}^{2}}\].
The length of the copper wire is ${{L}_{2}}=3.5m$.
Area of crosssection of the copper wire is \[{{A}_{2}}=4.0\times {{10}^{5}}{{m}^{2}}\] .
Now,
The change in length is given by:
$\Delta L={{L}_{1}}{{L}_{2}}=4.73.5=1.2m$
Let the force applied in both the cases be $F$ .
Therefore, Young’s modulus of the steel wire is given by:
${{Y}_{1}}=\frac{F}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta L}=\frac{F\times 4.7}{3.0\times {{10}^{5}}\times 1.2}$ …… (1)
And Young’s modulus of the copper wire is given by:
${{Y}_{2}}=\frac{F}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta L}=\frac{F\times 3.5}{4.0\times {{10}^{5}}\times 1.2}$ …… (2)
Therefore,
$\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{F\times 4.7\times 4.0\times {{10}^{5}}\times 1.2}{3.0\times {{10}^{5}}\times 1.2\times F\times 3.5}=\frac{1.79}{1}$
Hence the ratio of Young’s modulus of steel to that of copper is $1.79:1$ .
2. Figure shows the strainstress curve for a given material. What are
a) Young’s modulus?
Ans: From the graph given in the above question it is clear that:
Stress for a given material is $150\times {{10}^{6}}N/{{m}^{2}}$ and strain is $0.002$.
Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Y=\frac{150\times {{10}^{6}}N/{{m}^{2}}}{0.002}=7.5\times {{10}^{10}}N/{{m}^{2}}$
Therefore, Young’s modulus for the given material is $7.5\times {{10}^{10}}N/{{m}^{2}}$ .
b) Approximate yield strength for this material?
Ans: The yield strength of the material is the maximum stress it sustains without crossing the elastic limit.
From the graph given in the above question, it is clear that approximate yield strength for this material is $300\times {{10}^{6}}N/{{m}^{2}}$ or $3\times {{10}^{8}}N/{{m}^{2}}$ .
3. The stressstrain 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?
Ans: In the two graphs it is that given that stress for A is more than that of B.
As,
\[Young's\text{ }modulus=\frac{Stress}{Strain}\],
Therefore, material A has greater Young's modulus.
b) Which of the two is the stronger material?
Ans: The strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point.
Fracture point is defined as the extreme point in a stressstrain curve.
From the graph it is clear that material A can withstand more strain than material B.
Therefore, material A is stronger than material B.
4. 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.
Ans: The given statement is false.
As there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain. Therefore, the Young’s modulus of steel is greater than that of rubber.
b) The stretching of a coil is determined by its shear modulus.
Ans: The given statement is true.
As the shear modulus of a coil relates with the change in shape of the coil and the stretching of coil changes its shape without any change in the length. Therefore, the shear modulus of elasticity is involved. Hence the stretching of a coil is determined by its shear modulus.
5. 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.
Ans: In the above question it is given that:
Diameter of the wires is $d=0.25m$.
Hence $r=0.125m$.
Length of the steel wire is ${{L}_{1}}=1.5m$.
Length of the brass wire is ${{L}_{2}}=1.0m$.
Total force exerted on the steel wire is ${{F}_{1}}=\left( 4+6 \right)g=10g$
$\therefore {{F}_{1}}=10\times 9.8=98N$.
Young’s modulus for steel is given by
${{Y}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta {{L}_{1}}}$
Where,
$\Delta {{L}_{1}}$ is the change in the length of the steel wire.
And ${{A}_{1}}$ is the area of crosssection of the steel wire.
$\therefore {{A}_{1}}=\pi {{r}_{1}}^{2}$
We have,
Young’s modulus of steel is ${{Y}_{1}}=2.0\times {{10}^{11}}Pa$.
$\Rightarrow \Delta {{L}_{1}}=\frac{{{F}_{1}}\times {{L}_{1}}}{{{A}_{1}}\times {{Y}_{1}}}$
$\Rightarrow \Delta {{L}_{1}}=\frac{98\times 1.5}{\pi {{\left( 0.125 \right)}^{2}}\times 2.0\times {{10}^{11}}}=1.49\times {{10}^{4}}m$.
Total force on the brass wire is ${{F}_{2}}=6\times 9.8=58.8N$.
Young’s modulus for brass is given by
${{Y}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta {{L}_{2}}}$
Where,
$\Delta {{L}_{2}}$ is the change in the length of the brass wire.
And ${{A}_{2}}$ is the area of crosssection of the brass wire.
$\therefore {{A}_{2}}=\pi {{r}_{2}}^{2}$
We have,
Young’s modulus of brass is ${{Y}_{2}}=0.91\times {{10}^{11}}Pa$.
$\Rightarrow \Delta {{L}_{2}}=\frac{{{F}_{2}}\times {{L}_{2}}}{{{A}_{2}}\times {{Y}_{2}}}$
$\Rightarrow \Delta {{L}_{2}}=\frac{58.8\times 1}{\pi {{\left( 0.125 \right)}^{2}}\times 0.91\times {{10}^{11}}}=1.3\times {{10}^{4}}m$.
Clearly, the elongation of the steel wire is $1.49\times {{10}^{4}}m$ and that of the brass wire is $1.3\times {{10}^{4}}m$.
6. 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?
Ans: In the above question it is given that:
Edge of the aluminium cube is $L=10cm=0.1m$.
The mass attached to the cube is $m=100kg$.
Shear modulus $\eta $ of aluminium is $25GPa=25\times {{10}^{10}}Pa$.
We know that:
\[Shear\text{ }modulus\left( \eta \right)=\frac{\text{Shear stress}}{\text{Shear strain}}\]
$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{\left( \frac{L}{\Delta L} \right)}$
Where,
$F$ is the applied force.
$\therefore F=mg=100\times 9.8=980N$
Area of one of the faces of the cube is $A=0.1\times 0.1=0.01{{m}^{2}}$.
Vertical deflection of the cube is $\Delta L$.
$\Delta L=\frac{FL}{A\eta }$
$\Rightarrow \Delta L=\frac{980\times 0.1}{0.01\times 25\times {{10}^{9}}}=3.92\times {{10}^{7}}m$
Clearly, the vertical deflection of this face of the cube is $3.92\times {{10}^{7}}m$.
7. 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.
Ans: In the above question it is given that:
Mass of the big structure is $M=50000kg$.
Inner radius of the column is $r=30cm=0.3m$.
Outer radius of the column is $R=60cm=0.6m$
Young’s modulus of steel is $Y=2\times {{10}^{11}}Pa$
The total force exerted is $F=Mg=50000\times 9.8N$.
\[\text{Stress = Force exerted on a single column}\]
$\Rightarrow Stress=\frac{50000\times 9.8}{4}=122500N$
Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Strain=\frac{\left( \frac{F}{A} \right)}{Y}$
Where,
Area is given by
$A=\pi \left( {{R}^{2}}{{r}^{2}} \right)=\pi \left( {{\left( 0.6 \right)}^{2}}{{\left( 0.3 \right)}^{2}} \right)$.
$\Rightarrow Strain=\frac{\left( \frac{50000\times 9.8}{\pi \left( {{\left( 0.6 \right)}^{2}}{{\left( 0.3 \right)}^{2}} \right)} \right)}{2\times {{10}^{11}}}=7.22\times {{10}^{7}}$
Therefore, the compressional strain of each column is $7.22\times {{10}^{7}}$.
8. A piece of copper having a rectangular crosssection of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Ans: In the above question it is given that:
Length of the piece of copper is $l=19.1mm=19.1\times {{10}^{3}}m$.
Breadth of the piece of copper is $b=15.2mm=15.2\times {{10}^{3}}m$
Area of the copper piece will be:
$A=l\times b$
$\Rightarrow A=19.1\times {{10}^{3}}\times 15.2\times {{10}^{3}}=2.9\times {{10}^{4}}{{m}^{2}}$
Tension force applied on the piece of copper is $F=44500N$.
Modulus of elasticity of copper is $\eta =42\times {{10}^{9}}N/{{m}^{2}}$.
We know that :
\[\text{Modulus of elasticity}\left( \eta \right)=\frac{Stress}{Strain}\]
$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{Strain}$
$\Rightarrow Strain=\frac{F}{A\eta }$
$\Rightarrow Strain=\frac{44500}{2.9\times {{10}^{4}}\times 42\times {{10}^{9}}}=3.65\times {{10}^{3}}$
Hence, the resulting strain is $3.65\times {{10}^{3}}$.
9. 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 ${{10}^{8}}N/{{m}^{2}}$, what is the maximum load the cable can support?
Ans: In the above question it is given that:
Radius of the steel cable is $r=1.5cm=0.015m$.
Maximum allowable stress is ${{10}^{8}}N/{{m}^{2}}$.
We know that:
\[\text{Maximum force = Maximum stress }\times \text{Area of crosssection}\]
\[\Rightarrow \text{Maximum force =1}{{\text{0}}^{8}}\times \pi {{\left( 0.015 \right)}^{2}}=7.065\times {{10}^{4}}N\]
Therefore, the cable can support the maximum load of \[7.065\times {{10}^{4}}N\].
10. 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.
Ans: In the above question it is given that:
Tension force acting on each wire is the same.
Therefore the extension produced in each wire is the same.
As the length of both wires is the same, the strain in both wires is also the same.
Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{Strain}=\frac{\frac{4F}{\pi {{d}^{2}}}}{Strain}$ …… (1)
Where,
$F$ is the Tension force,
$A$ is the area of crosssection and
$d$ is the diameter of the wire
From equation (1), it is clear that $Y\propto \frac{1}{{{d}^{2}}}$.
Young’s modulus for iron is ${{Y}_{1}}=190\times {{10}^{9}}Pa$.
Let the diameter of the iron wire be ${{d}_{1}}$.
Young’s modulus for copper is ${{Y}_{2}}=100\times {{10}^{9}}Pa$.
Let the diameter of the copper wire be ${{d}_{2}}$.
Therefore, the ratio of their diameters is given as:
$\frac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\frac{{{Y}_{1}}}{{{Y}_{2}}}}=\sqrt{\frac{190\times {{10}^{9}}}{100\times {{10}^{9}}}}=\frac{1.31}{1}$
Therefore, the ratio of diameters of copper wire to iron wire is $1.31:1$.
11. 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 \[0.065c{{m}^{2}}\]. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Ans: In the above question it is given that:
Mass is \[m=14.5\text{ }kg\].
Length of the steel wire is \[l=1.0\text{ }m\]
Angular velocity is $\omega =2rev/s$
Crosssectional area of the wire is \[a=0.065c{{m}^{2}}=0.065\times {{10}^{4}}{{m}^{2}}\].
Consider the elongation of the wire when the mass is at the lowest point of its path to be $\Delta l$.
The total force on the mass when the mass is placed at the position of the vertical circle is given by:
$F=mg+ml{{\omega }^{2}}$
$\Rightarrow F=14.5\times 9.8+14.5\times 1\times {{2}^{4}}=200.1N$
Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{\left( \frac{\Delta l}{l} \right)}$
$\Rightarrow \Delta l=\frac{Fl}{AY}$
We know that Young’s modulus for steel is $2\times {{10}^{11}}Pa$.
Therefore,
$\Delta l=\frac{220.1\times 1}{0.065\times {{10}^{4}}\times 2\times {{10}^{11}}}=1.53\times {{10}^{4}}m$
Thus, the elongation of the wire is $1.53\times {{10}^{4}}m$.
12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm$\left( 1atm=1.013\times {{10}^{5}}Pa \right)$. 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.
Ans: In the above question it is given that:
Initial volume is ${{V}_{1}}=100.0l=100\times {{10}^{3}}{{m}^{3}}$.
Final volume is ${{V}_{2}}=100.5l=100.5\times {{10}^{3}}{{m}^{3}}$.
Thus, the increase in volume is ${{V}_{2}}{{V}_{1}}=0.5\times {{10}^{3}}{{m}^{3}}$.
Increase in pressure is $\Delta p=100atm=100\times 1.013\times {{10}^{5}}Pa$.
The formula for bulk modulus is
\[\text{Bulk Modulus}=\frac{\Delta p}{\left( \frac{\Delta V}{{{V}_{1}}} \right)}=\frac{\Delta p{{V}_{1}}}{\Delta V}\]
\[\Rightarrow \text{Bulk Modulus=}\frac{100\times 1.013\times {{10}^{5}}\times 100\times {{10}^{3}}}{0.5\times {{10}^{3}}{{m}^{3}}}=2.026\times {{10}^{9}}Pa\]
We know that Bulk modulus of air is $1\times {{10}^{5}}Pa$.
$\Rightarrow \frac{Bulk\text{ }modulus\text{ }of\text{ }water}{Bulk\text{ }modulus\text{ }of\text{ }air}=\frac{2.026\times {{10}^{9}}}{1\times {{10}^{5}}}=2.026\times {{10}^{4}}$
This ratio is very high because air is more compressible than water.
13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times {{10}^{3}}kg/{{m}^{3}}$ ?
Ans: In the above question it is given that:
Pressure at the given depth is $p=80.0atm=80\times 1.01\times {{10}^{5}}Pa$.
Consider the given depth to be h.
Density of water at the surface is ${{\rho }_{1}}=1.03\times {{10}^{3}}kg/{{m}^{3}}$
Consider ${{\rho }_{2}}$ to be the density of water at the depth h.
Consider ${{V}_{1}}$ to be the volume of water of mass m at the surface.
Consider ${{V}_{2}}$ to be the volume of water of mass m at the depth h.
Consider $\Delta V$ to be the change in volume.
$\Delta V={{V}_{1}}{{V}_{2}}$
$\Rightarrow \Delta V=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)\left( \frac{1}{{{\rho }_{2}}} \right) \right]$
Now,
$Volumetric\text{ }strain=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)\left( \frac{1}{{{\rho }_{2}}} \right) \right]\times \left( \frac{{{\rho }_{1}}}{m} \right)$
$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=1\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)$ …… (1)
Bulk modulus is given by:
\[Bulk\text{ }modulus=\frac{p{{V}_{1}}}{\Delta V}\]
$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=\frac{p}{B}$
Compressibility of water is given by:
$\frac{1}{B}=45.8\times {{10}^{11}}P{{a}^{1}}$
\[\Rightarrow \frac{\Delta V}{{{V}_{1}}}=80\times 1.013\times {{10}^{5}}\times 45.8\times {{10}^{11}}=3.71\times {{10}^{3}}\] …… (2)
From equations (1) and (2) we get:
$1\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)=3.71\times {{10}^{3}}$
$\Rightarrow {{\rho }_{2}}=\frac{1.03\times {{10}^{3}}}{\left[ 1\left( 3.71\times {{10}^{3}} \right) \right]}=1.034\times {{10}^{3}}kg{{m}^{3}}$
Clearly, the density of water at the given depth (h) is $1.034\times {{10}^{3}}kg{{m}^{3}}$.
14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Ans: In the above question it is given that:
The hydraulic pressure exerted on the glass slab is $p=10atm=10\times 1.013\times {{10}^{5}}Pa$.
Also, we know that the bulk modulus of glass is $B=37\times {{10}^{9}}N/{{m}^{2}}$.
Bulk modulus is given by the relation:
\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]
Where,
\[\frac{\Delta V}{V}\] is the fractional change in volume.
\[\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{p}{B}\]
\[\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{10\times 1.013\times {{10}^{5}}}{37\times {{10}^{9}}}=2.73\times {{10}^{5}}\]
Clearly, the fractional change in the volume of the glass slab is \[2.73\times {{10}^{5}}\].
15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7\times {{10}^{6}}Pa$.
Ans: In the above question it is given that:
The length of an edge of the solid copper cube is $l=10cm=0.1m$.
Hydraulic pressure is $p=7\times {{10}^{6}}Pa$.
Bulk modulus of copper is $B=140\times {{10}^{9}}Pa$.
Bulk modulus is given by the relation:
\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]
Where,
\[\frac{\Delta V}{V}\] is the volumetric strain
$\Delta V$ is the change in volume.
$V$ is the original volume.
\[\Rightarrow \Delta V=\frac{pV}{B}\]
The original volume of the cube is $V={{l}^{3}}$.
\[\Rightarrow \Delta V=\frac{p{{l}^{3}}}{B}\]
\[\Rightarrow \Delta V=\frac{7\times {{10}^{6}}\times {{\left( 0.1 \right)}^{3}}}{140\times {{10}^{9}}}=5\times {{10}^{8}}{{m}^{3}}=5\times {{10}^{2}}c{{m}^{3}}\]
Clearly, the volume contraction of the solid copper cube is \[5\times {{10}^{2}}c{{m}^{3}}\].
16. How much should the pressure on a litre of water be changed to compress it by 0.10%?
Ans: In the above question it is given that:
Volume of water is $V=1L$.
The water is to be compressed by \[0.10%\].
$\therefore Fractional\text{ }change=\frac{\Delta V}{V}=\frac{0.1}{100\times 1}={{10}^{3}}$
Bulk modulus is given by the relation:
\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]
\[\Rightarrow p=B\times \left( \frac{\Delta V}{V} \right)\]
We know that, bulk modulus of water is $B=2.2\times {{10}^{9}}N/{{m}^{2}}$
\[\Rightarrow p=2.2\times {{10}^{9}}\times {{10}^{3}}=2.2\times {{10}^{6}}N/{{m}^{2}}\].
Clearly, the pressure on water should be \[2.2\times {{10}^{6}}N/{{m}^{2}}\].
17. Anvils made of single crystals of diamond, with the shape as shown in figure, 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?
Ans: In the above question it is given that:
The diameter of the cones at the narrow ends is $d=0.50mm=0.5\times {{10}^{3}}m$
$\therefore $Radius will be $r=0.25\times {{10}^{3}}m$
The Compressional force is \[F=50000N\].
Pressure at the tip of the anvil is given by:
\[P\text{ }=\text{ }\frac{Force}{Area}=\frac{50000}{\pi {{\left( 0.25\times {{10}^{3}} \right)}^{2}}}=2.55\times {{10}^{11}}Pa\]
Clearly, the pressure at the tip of the anvil will be \[2.55\times {{10}^{11}}Pa\].
18. 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 crosssectional areas of wires A and B are $1.0m{{m}^{2}}$ and $2.0m{{m}^{2}}$, respectively. At what point along the rod should a mass m be suspended in order to produce
a) Equal stresses
Ans: In the above question it is given that:
The crosssectional area of wire A is $1.0m{{m}^{2}}=1.0\times {{10}^{6}}{{m}^{2}}$.
Crosssectional area of wire B is $2.0m{{m}^{2}}=2.0\times {{10}^{6}}{{m}^{2}}$.
Young’s modulus for steel is ${{Y}_{1}}=2\times {{10}^{11}}N/{{m}^{2}}$ .
Young’s modulus for aluminium is ${{Y}_{2}}=7\times {{10}^{10}}N/{{m}^{2}}$.
Consider a small mass m to be suspended to the rod at a distance y from the end where wire A is attached.
\[Stress\text{ }in\text{ }the\text{ }wire=\frac{F}{a}\]
If the two wires have equal stresses,
\[\frac{{{F}_{1}}}{{{a}_{1}}}=\frac{{{F}_{2}}}{{{a}_{2}}}\]
Where,
${{F}_{1}}$ is the force exerted on the steel wire.
${{F}_{2}}$ is the force exerted on the aluminium wire.
\[\therefore \frac{{{F}_{1}}}{{{a}_{1}}}=\frac{{{F}_{2}}}{{{a}_{2}}}=\frac{1}{2}\] …… (1)
Consider the figure given below:
Taking torque about the point of suspension, we get:
${{F}_{1}}y={{F}_{2}}\left( 1.05y \right)$ …… (2)
From equations (1) and (2), we get:
$\left( 1.05y \right)=\frac{1}{2}$
$\Rightarrow 2\left( 1.05y \right)=y$
$\Rightarrow y=0.7m$
Clearly, to produce an equal stress in the two wires, the mass must be suspended at a distance of $0.7m$ from the end where wire A is attached.
b) Equal strains in both steel and aluminium wires.
Ans: Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Strain=\frac{Stress}{Y}=\frac{\left( \frac{F}{a} \right)}{Y}$
When the strain in the two wires is equal,
$\frac{\left( \frac{{{F}_{1}}}{{{a}_{1}}} \right)}{{{Y}_{1}}}=\frac{\left( \frac{{{F}_{2}}}{{{a}_{2}}} \right)}{{{Y}_{2}}}$
$\Rightarrow \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{a}_{1}}{{Y}_{1}}}{{{a}_{2}}{{Y}_{2}}}$
We have
$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}$
$\Rightarrow \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{1\times 2\times {{10}^{11}}}{2\times 7\times {{10}^{10}}}=\frac{10}{7}$ …… (3)
Consider the torques about the point where mass m, to be suspended at a distance ${{y}_{1}}$ from the side where wire A attached;
${{F}_{1}}{{y}_{1}}={{F}_{2}}\left( 1.05{{y}_{1}} \right)$
\[\Rightarrow \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{\left( 1.05{{y}_{1}} \right)}{{{y}_{1}}}\] …… (4)
From equations (3) and (4), we get:
$\left( 1.05{{y}_{1}} \right)=\frac{10}{7}$
$\Rightarrow 7\left( 1.05{{y}_{1}} \right)=10{{y}_{1}}$
$\Rightarrow {{y}_{1}}=0.432m$
Clearly, to produce an equal strain in the two wires, the mass must be suspended at a distance of $0.432m$from the end where wire A is attached.
19. A mild steel wire of length 1.0 m and crosssectional area $0.50\times {{10}^{2}}c{{m}^{2}}$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the midpoint of the wire. Calculate the depression at the midpoint.
Ans: In the above question it is given that:
Mass is \[m=100\text{ }g=0.100\text{ k}g\]
Length of the mild steel wire is $1.0m$.
Crosssectional area is $0.50\times {{10}^{2}}c{{m}^{2}}$.
Now consider the figure given below:
If $x$ is the depression at the midpoint i.e., $CD=x$.
$AB=BC=l=0.5m$
$AD=BD=\sqrt{{{l}^{2}}+{{x}^{2}}}$
The increase in length will be:
$\Delta l=AD+BDAB=2ADAB$
$\Rightarrow \Delta l=2\sqrt{{{l}^{2}}+{{x}^{2}}}2l$
$\Rightarrow \Delta l=2l\left( 1+\frac{{{x}^{2}}}{2{{l}^{2}}} \right)2l$
$\Rightarrow \Delta l=2l\left( 1+\frac{{{x}^{2}}}{2{{l}^{2}}}1 \right)=\frac{{{x}^{2}}}{l}$
Now, we know that
$\text{Strain=}\frac{\text{increase in length}}{\text{original length}}=\frac{{{x}^{2}}}{2{{l}^{2}}}$
If T is the tension in the wire, then
$2T\cos \theta =mg$
$\Rightarrow T=\frac{mg}{2\cos \theta }\to (1)$
Here,
$\cos \theta =\frac{x}{{{\left( {{l}^{2}}+{{x}^{2}} \right)}^{\frac{1}{2}}}}=\frac{x}{l{{\left( 1+\frac{{{x}^{2}}}{{{l}^{2}}} \right)}^{\frac{1}{2}}}}=\frac{x}{l\left( 1+\frac{{{x}^{2}}}{2{{l}^{2}}} \right)}$
As $x<<l$;
$1+\frac{{{x}^{2}}}{2{{l}^{2}}}\approx 1$
$\Rightarrow \cos \theta =\frac{x}{l}\to (2)$
From (1) and (2),
$T=\frac{mgl}{2x}$
Also, stress is given by:
\[Stress=\frac{T}{A}=\frac{mgl}{2Ax}\]
And Young’s modulus is given by:
$Y=\frac{Stress}{Strain}$
$\Rightarrow Y=\frac{mgl}{2Ax}\times \frac{2{{l}^{2}}}{{{x}^{2}}}=\frac{mg{{l}^{3}}}{A{{x}^{3}}}$
$\Rightarrow x=l{{\left( \frac{mg}{AY} \right)}^{\frac{1}{3}}}$
$\Rightarrow x=0.5{{\left[ \frac{0.1\times 10}{20\times {{10}^{11}}\times 0.5\times {{10}^{6}}} \right]}^{\frac{1}{3}}}=0.01074m$
Clearly, the depression at the midpoint is $0.01074m$.
20. 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 $6.9\times {{10}^{7}}Pa$ ? Assume that each rivet is to carry one quarter of the load.
Ans: In the above question it is given that:
The diameter of the metal strip is $d=6.0mm=6.0\times {{10}^{3}}m$.
Radius will be $r=3.0\times {{10}^{3}}m$.
Maximum shearing stress is $6.9\times {{10}^{7}}Pa$.
We know that:
\[Maximum\text{ }stress\text{ }=\text{ }\frac{Maximum\text{ }load\text{ }or\text{ }force}{Area}\]
\[\Rightarrow Maximum\text{ }force\text{ }=\text{ }Maximum\text{ }stress\text{ }\times \text{ }Area\]
\[\Rightarrow Maximum\text{ }force\text{ }=\text{ }6.9\times {{10}^{7}}\times \pi {{\left( r \right)}^{2}}\]
\[\Rightarrow Maximum\text{ }force\text{ }=\text{ }6.9\times {{10}^{7}}\times \pi {{\left( 3.0\times {{10}^{3}} \right)}^{2}}=1949.94N\]
Since each rivet is said to carry one quarter of the load;
Maximum tension on each rivet is \[1949.94\times 4=7799.76N\]
Clearly, the maximum tension that can be exerted is \[7799.76N\].
21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about $1.1\times {{10}^{8}}Pa$ . A steel ball of initial volume is $0.32{{m}^{3}}$ dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom?
Ans: In the above question it is given that:
Water pressure at the bottom of the trench is about $1.1\times {{10}^{8}}Pa$.
Initial volume of the steel ball is $V=0.32{{m}^{3}}$.
Bulk modulus of steel is $1.6\times {{10}^{11}}N/{{m}^{2}}$.
The ball is said to fall at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Consider the change in the volume of the ball on reaching the bottom of the trench to be $\Delta V$.
Bulk modulus is given by the relation:
\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]
\[\Rightarrow \Delta V=\frac{pV}{B}\]
\[\Rightarrow \Delta V=\frac{1.1\times {{10}^{8}}\times 0.32}{1.6\times {{10}^{11}}}=2.2\times {{10}^{4}}{{m}^{3}}\]
Clearly, the change in volume of the ball on reaching the bottom of the trench is \[2.2\times {{10}^{4}}{{m}^{3}}\].
NCERT Solution of Physics Class 11 Chapter 9  Free PDF Download
The subject of Physics can be quite demanding, and students often struggle with it. The curated solutions answer all your Physics Class 11 NCERT solutions Chapter 9 questions. It is an excellent way for students to improve their learning. If you wish to get a better grip on the concepts of Physics, our NCERT Physics Class 11 Chapter 9 solutions are perfect for you.
Mechanical Properties of Solids Class 11 NCERT PDF is now available for you to download. The solution is in a PDF format for your convenience and easy accessibility. You can download it for free from the Vedantu App.
NCERT Solutions for Class 11 Physics Chapter 9 PDF Topics
Solids have some characteristic responses when exposed to an applied force. NCERT solutions for Class 11 Physics Chapter 9 provide indepth knowledge on the mechanical properties of solids. It covers some vital topics, like:
Elastic Behaviour of Solids
When force is applied to solids, they undergo deformation. However, the material is elastic which will allow it to return to its initial form when the force is withdrawn.
Stress and Strain
Stress is the force applied to per unit area of a material, whereas strain is the extension per unit length of the said material. It is explained in detail in NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids.
Hooke’s Law
English Scientist Robert Hooke states that the size of deformation of an object is directly proportional to the force of deformation.
Stress  Strain Curve
The stress and strain values of a material under differing loads can be plotted on a graph, thereby giving us a stressstrain curve.
Elastic Moduli
Most solids resist elastic deformation when stress is applied. Elastic modulus is the quantity that measures this resistance.
Application of Elastic Behaviour of Materials
The theory of elasticity of materials is effectively put to practical uses, such as architecture and construction.
Studying these topics can be made easy by referring to the study materials. Class 11 Physics Mechanical Properties of Solids.
NCERT Solutions help you understand the basics; a strong understanding of the subject will help you master the tougher topics of Physics.
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What are Mechanical Properties?
Every material reacts to applied force. When force is applied to a material, it shows some physical properties. These properties are called mechanical properties. NCERT solutions for Class 11 Physics Mechanical Properties of Solids will further provide an insight on the topic.
Most substances are anisotropic  meaning that the properties of material vary with orientation. This diversity in properties helps us classify various substances into various groups based on their similarity as well as differences. Ch 9 Physics Class 11 NCERT solutions try to establish the identity of various materials based on their reaction to an applied force.
The most common mechanical properties are:
Strength
Ductility
Fracture toughness
Hardness
Impact resistance
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FAQs on NCERT Solutions for Class 11 Physics Chapter 9  Mechanical Properties Of Solids
1. What are Some Mechanical Properties of Solids?
Mechanical Properties of Solids Class 11 NCERT PDF go into the depths of the topic. It also describes the various mechanical properties of solids, some of which are as follows:
Brittleness: Brittleness is the property of solids due to which they break into smaller pieces when enough force is applied.
Compressive Strength: Due to their compressive strength, solids resist being pushed together.
Ductility: It is the property of solids that results in their deformation under tensile strength.
Elasticity: An elastic substance is capable of resisting distortions or deformations under an applied force. These materials return to their original shape when the force is removed.
Other than that, solids also show plasticity, which is the ability to undergo permanent deformation when a force is applied.
2. What is Elastic Moduli?
Most solids resist elastic deformation when stress is applied. Elastic modulus is the quantity that measures this resistance. It is an important inclusion of Class 11 Physics Chapter 9.
3. What is Hooke’s Law?
Hooke’s law states that the size of deformation of an object is directly proportional to the force of deformation.
4. What is an intermolecular force?
As discussed in the chapter, the force of attraction or repulsion acting between the particles (atoms, molecules and ions) neighbouring each other, is referred to as the intermolecular force of attraction or repulsion. Intermolecular forces are considered weaker in comparison to the intramolecular (covalent and ionic bonds) force acting between the atoms and molecules. Examples of intermolecular force are the London dispersion force, iondipole interaction etc.
5. What are the practical applications of the Mechanical Properties of Solids Class 11?
Solid mechanics has a plethora of reallife applications. It is one of the most fundamental applied engineering sciences which plays a crucial role in explaining a lot of physical phenomena around us and in explaining human anatomy. It is also used extensively in surgical implants and the design of dental prostheses. The principles and properties of solid mechanics have also been used in the Euler–Bernoulli beam equation.
6. Where can I get the NCERT Solutions for Class 11 Physics, Chapter 9?
To avail the NCERT Solutions for Class 11 Physics, Chapter 9, refer to Vedantu's NCERT Solutions for this chapter. After clicking on the given link, a web page will open, click on the download option to download the PDF of Vedantu's NCERT Solutions for this chapter or you can also continue your studies online. It is available free of cost and can be assessed online anytime, anywhere. These solutions will clear all your conceptual doubts and will help you to fetch good marks in this Physics chapter.
7. What do I do to score well in Class 11 Physics, Chapter 9?
First of all, to study Class 11 Physics, Chapter 9, you must carefully read from the NCERT textbook to understand all basic concepts well. Once you have learnt and understood the chapter basics, proceed to the NCERT back exercise questions for this chapter. If you get stuck in between, refer to Vedantu's NCERT Solutions for this chapter to clear all doubts as soon as possible. Last but not the least, practice additional questions available on the Vedantu app and website and consistently revise this chapter to fetch a high score in the Physics exam.
8. Is Class 11 Physics Chapter 9 difficult?
Class 11 Physics, Chapter 9Mechanical Properties of Solids can be made easy if you follow the right strategy and approach while preparing this chapter. After studying this chapter from the NCERT and solving the back exercise questions, you should solve the previous year questions from this chapter in the Physics exam. Focus more on numerical problems and prepare a list of all the formulae from this chapter. Through consistent practice and optimism, you too can master this chapter.