# NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids: If a rubber band is stretched by applying a variable force after a particular amount of force the rubber band brakes. The breakage happens when the applied force is beyond the elastic limit. To compare the elastic limit we use the term modulus of elasticity. The solutions of NCERT class 11 physics chapter 9 mechanical properties of solids mainly deals with questions related to the concepts of stress, strain, and modulus of elasticity. If these concepts are clear then it is very easy to understand the CBSE NCERT solutions for class 11 physics chapter 9 mechanical properties of solids. The chapter is small in length and easy to score. NCERT solutions help students in their preparations.

Exercise

## Below are the important formulas required for NCERT solutions for class 11 physics chapter 9 mechanical properties of solids

• $Stress=\frac{Force}{Area}$

• Hooke's law

• $E=\frac{Stress }{Strain}$

$E= modulus \: of \: elasticity$

• longitudinal strain

The ratio of change in length to the original length.

• A wire of length L and radius r is clamped to a rigid support and a mass m is attached to the other end, then Young's modulus is

$Y=\frac{mgL}{\pi r^{2}\Delta l}$

L = Original length

$\Delta l$  = Change in length

## NCERT solutions for class 11 physics chapter 9 mechanical properties of solids exercise

Let the Young's Modulus of steel and copper be YS and YC respectively.

Length of the steel wire l= 4.7 m

Length of the copper wire l= 4.7 m

The cross-sectional area of the steel wire AS = $3.0 \times 10^{-5} m^2$

The cross-sectional area of the Copper wire AC = $4.0 \times 10^{-5} m^2$

Let the load and the change in the length be F and $\Delta l$ respectively

$\\Y=\frac{Fl}{A\Delta l}\\ \frac{F}{\Delta l}=\frac{AY}{l}$

Since F and $\Delta l$ is the same for both wires we have

$\\\frac{A_{S}Y_{S}}{l_{S}}=\frac{A_{C}Y_{C}}{l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{A_{C}l_{S}}{A_{S}l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{4.0\times 10^{-5}\times 4.7}{3.0\times 10^{-5}\times3.5}\\ \frac{Y_{S}}{Y_{C}}=1.79$

The ratio of Young’s modulus of steel to that of copper is 1.79.

Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.

For the given material

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

The Yield Strength is approxmiately $3\times 10^{8}Nm^{-2}$ for the given material. We can see above this value of strain, the body stops behaving elastically.

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.

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.

The Young’s modulus of rubber is greater than that of steel;

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.

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.

Tension in the steel wire is F1

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

Length of steel wire l1 = 1.5 m

The diameter of the steel wire, d = 0.25 cm

$\\A=\pi \left ( \frac{d}{2} \right )^{2}\\ A=\pi \times \left ( \frac{0.25\times 10^{-2}}{2} \right )^{2}\\ A=4.9\times 10^{-6}m^{2}$

Area od the steel wire, $A=4.9\times 10^{-6}\ m^{2}$

Let the elongation in the steel wire be $\Delta l_{1}$

Young's Modulus of steel, Y1 = $2\times 10^{11}Nm^{-2}$

$\\Y_{1}=\frac{F_{1}l_{1}}{\Delta l_{1}A }\\ \Delta l_{1}=\frac{F_{1}l_{1}}{Y_{1}A}\\ \Delta l_{1}=\frac{98\times 1.5}{2\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{1}=1.5\times 10^{-4}\ m$

Tension in the Brass wire is F2

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

Length of Brass wire l2 = 1.5 m

Area od the brass wire, $A=4.9\times 10^{-6}\ m^{2}$

Let the elongation in the steel wire be $\Delta l_{2}$

Young's Modulus of steel, Y2 = $0.91\times 10^{11}Nm^{-2}$

$\\ \Delta l_{2}=\frac{F_{2}l_{2}}{Y_{2}A}\\ \Delta l_{2}=\frac{58.8\times 1}{0.91\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{2}=1.32\times 10^{-4}\ m$

Edge of the aluminium cube, l = 10 cm = 0.1 m

Area of a face of the Aluminium cube, A = l2 = 0.01 m2

Tangential Force is F

$\\F=100\times 9.8\\ F=980\ N$

Tangential Stress is F/A

$\\\frac{F}{A}=\frac{980}{0.01}\\ \frac{F}{A}=98000N$

Shear modulus of aluminium $\eta =2.5\times 10^{10}Nm^{-2}$

$\\Tangential\ Strain=\frac{\frac{F}{A}}{\eta }\\ =\frac{98000}{2.5\times 10^{10}}\\ =3.92\times 10^{-6}$

Let the Vertical deflection be $\Delta l$

$\\\Delta l=Tangential\ Stress\times Side \\\Delta l=3.92\times 10^{-6}\times 0.1\\ \Delta l=3.92\times 10^{-7}m$

Inner radii of each column, r1 = 30 cm = 0.3 m

Outer radii of each colum, r2 = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

$\\P=\frac{50000\times 9.8}{4\times \pi \times (0.6^{2}-0.3^{2})}\\ P=1.444\times 10^{5}Nm^{-2}$

Youngs Modulus of steel is $Y=2\times 10^{11}Nm^{-2}$

$\\Compressional\ strain=\frac{P}{Y} \\=\frac{1.444\times 10^{5}}{2\times 10^{11}}\\ =7.22\times 10^{-7}$

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, $\eta =42\times 10^{9}Nm^{-2}$

$\\Strain=\frac{F}{A\eta }\\ =\frac{F}{lb\eta }\\ =\frac{44500}{15.2\times 10^{-3}\times 19.1\times 10^{-3}\times 42\times 10^{9}}\\ =3.65\times 10^{-3}$

Let the maximum Load the Cable Can support be T

Maximum Stress Allowed, P = 108 N m-2

Radius of Cable, r = 1.5 cm

$\\T=P\pi r^{2}\\ T=10^{8}\times \pi \times (1.5\times 10^{-2})^{2}\\ T=7.068\times 10^{4}N$

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.

$\\Y=\frac{Fl}{\Delta lA}\\ Y=\frac{4Fl}{\Delta l\pi d^{2}}\\ Yd^{2}=k$                                        As F, l and $\Delta l$ are equal for all wires

$Y_{iron}=1.9\times 10^{11}Nm^{-2}$

$Y_{copper}=1.1\times 10^{11}Nm^{-2}$

$\\\frac{d_{iron}}{d_{copper}}=\sqrt{\frac{Y_{copper}}{Y_{iron}}}\\ \frac{d_{iron}}{d_{copper}}=\sqrt{\frac{1.1\times 10^{11}}{1.9\times 10^{11}}}\\ \frac{d_{iron}}{d_{copper}}=0.761$

Mass of the body = 14.5 kg

Angular velocity, $\omega$ = 2 rev/s

$\omega =4\pi \ rad/s$

The radius of the circle, r = 1.0 m

Tension in the wire when the body is at the lowest point is T

$\\T=mg+m\omega ^{2}r\\ T=14.5\times 9.8+14.5\times (4\pi )^{2}\\ T=2431.84N$

Cross-Sectional Area of wire, A = 0.065 cm2

Young's Modulus of steel, $Y=2\times 10^{11}Nm^{-2}$

$\\\Delta l=\frac{Fl}{AY}\\ \Delta l=\frac{2431.84\times1}{0.065\times 10^{-4}\times 2\times 10^{11}}\\ \Delta l=1.87 mm$

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

$\\B=\frac{Stress}{Volumetric\ Strain}\\ B=\frac{P}{\frac{\Delta V}{V}}\\ B=\frac{100\times 1.013\times 10^{5}\times 100\times 10^{-3}}{0.5\times 10^{-3}}\\ B=2.026\times 10^{9}Nm^{-2}$

The bulk modulus of air is $B_{a}=1.0\times 10^{5}Nm^{-2}$

The Ratio of the Bulk Modulus of water to that of air is

$\\\frac{B}{B_{a}}=\frac{2.026\times 10^{9}}{1.0\times 10^{5}}\\ \frac{B}{B_{a}}=20260$

This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water.

Water at the surface is under 1 atm pressure.

At the depth, the pressure is 80 atm.

Change in pressure is $\Delta P=79\ atm$

Bulk Modulus of water is $B=2.2\times 10^{9}Nm^{-2}$

$\\B=-\frac{P}{\frac{\Delta V}{V}}\\ \frac{\Delta V}{V}=-\frac{P}{B} \\\frac{\Delta V}{V}=-\frac{79\times 1.013\times 10^{5}}{2.2\times 10^{9}}\\ \frac{\Delta V}{V}=-3.638\times 10^{-3}$

The negative sign signifies that for the same given mass the Volume has decreased

The density of water at the surface $\rho=1.03\times 10^{3}\ kg\ m^{-3}$

Let the density at the given depth be $\rho '$

Let a certain mass occupy V volume at the surface

$\\\rho=\frac{m}{V}\\$

$\\\rho'=\frac{m}{V+\Delta V}\\$

Dividing the numerator and denominator of RHS by V we get

$\\\rho'=\frac{\frac{m}{V}}{1+\frac{\Delta V}{V}}\\ \rho'=\frac{\rho }{1+\frac{\Delta V}{V}}\\ \rho'=\frac{1.03\times 10^{3}}{1+(-3.638\times 10^{-3})}\\ \rho'=1.034\times 10^{3}\ kg\ m^{-3}$

The density of water at a depth where pressure is 80.0 atm is $1.034\times 10^{3}\ kg\ m^{-3}$.

Bulk's Modulus of Glass is $B_{G}=3.7\times 10^{10}Nm^{-2}$

Pressure is P = 10 atm.

The fractional change in Volume would be given as

$\\\frac{\Delta V}{V}=\frac{P}{B}\\ \frac{\Delta V}{V}=\frac{10\times 1.013\times 10^{5}}{3.7\times 10^{10}}\\ \frac{\Delta V}{V}=2.737\times 10^{-5}$

The fractional change in Volume is $2.737\times 10^{-5}$

The bulk modulus of copper is $B_{C}=140GPa=1.4\times 10^{11}Nm^{-2}$

Edge of copper cube is s = 10 cm = 0.1 m

Volume Of copper cube is V = s3

V = (0.1)3

V = 0.001 m3

Hydraulic Pressure applies is $P=7.0\times 10^{6}Pa$

From the definition of bulk modulus

$\\B_{C}=\frac{P}{\frac{\Delta V}{V}}\\ \frac{\Delta V}{V}=\frac{P}{B_{C}}\\ \frac{\Delta V}{V}=\frac{7\times 10^{6}}{1.4\times 10^{11}} \frac{\Delta V}{V}=5\times 10^{-5}$

The volumetric strain is $5\times 10^{-5}$

Volume contraction will be

$\\\Delta V=Volumetric\ Strain\times Initial\ Volume\\ \Delta V=5\times 10^{-5}\times 10^{-3}\\ \Delta V=5\times 10^{-8}m^{3}\\ \Delta V=5\times 10^{-2}cm^{3}$

The volume contraction has such a small value even under high pressure because of the extremely large value of bulk modulus of copper.

Change in volume is $\Delta V=0.10 \%$

$\\Volumetric\ Strain =\frac{0.1}{100}\\ \frac{\Delta V}{V}=0.001$

Bulk modulus of water is $B_{w}=2.2\times 10^{9}\ Nm^{-2}$

$\\B_{w}=\frac{P}{\frac{\Delta V}{V}}\\ P=\frac{\Delta V}{V}\times B_{w}\\ P=0.001\times 2.2\times 10^{9}\\ P=2.2\times 10^{6}Pa$

A pressure of $2.2\times 10^{6}Pa$ is to be applied so that a litre of water compresses by 0.1%.

Note: The answer is independent of the volume of water taken into consideration. It only depends upon the percentage change.

## Q17  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?

The diameter of at the end of the anvil, d = 0.50 mm

Cross-sectional area at the end of the anvil is A

$\\A=\pi \left ( \frac{d}{2} \right )^{2}\\ A=\pi \times \left ( \frac{0.50\times 10^{-3}}{2} \right )^{2}\\ A=1.96\times 10^{-7}m^{2}$

Compressional Force applied, F = 50000N

The pressure at the tip of the anvil is P

$\\P=\frac{F}{A}\\ P=\frac{50000}{1.96\times 10^{-7}}\\ P=2.55\times 10^{11}Nm^{-2}$

The pressure at the tip of the anvil is $2.55\times 10^{11}Nm^{-2}$.

## Q18 (a)  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  equal stresses

Cross-Sectional Area of wire A is AA = 1 mm2

Cross-Sectional Area of wire B is AB = 2 mm2

Let the Mass m be suspended at x distance From the wire A

Let the Tension in the two wires A and B be FA and FB respectively

Since the Stress in the wires is equal

$\\\frac{F_{A}}{A_{A}}=\frac{F_{B}}{A_{B}}\\ \frac{F_{A}}{1}=\frac{F_{B}}{2}\\ 2F_{A}=F_{B}$

Equating moments of the Tension in the wires about the point where mass m is suspended we have

$\\xF_{A}=(1.05-x)F_{B}\\ xF_{A}=(1.05-x)\times 2F_{A}\\ x=2.1-2x\\ x=0.7m$

The Load should be suspended at a point 70 cm from a wire A such that there are equal stresses in the two wires.

Cross-Sectional Area of wire A is AA = 1 mm2

Cross-Sectional Area of wire B is AB = 2 mm2

Let the Mass m be suspended at y distance From the wire A

Let the Tension in the two wires A and B be FA and FB respectively

Since the Strain in the wires is equal

$\\\frac{F_{A}}{A_{A}Y_{S}}=\frac{F_{B}}{A_{B}Y_{Al}}\\ \frac{F_{A}}{F_{B}}=\frac{A_{A}Y_{S}}{A_{B}Y_{Al}}\\ \frac{F_{A}}{F_{B}}=\frac{1\times 2\times 10^{11}}{2\times 7\times 10^{10}}\\ \frac{F_{A}}{F_{B}}=\frac{10}{7}$

Equating moments of the Tension in the wires about the point where mass m is suspended we have

$\\yF_{A}=(1.05-y)F_{B}\\ yF_{A}=(1.05-y)\times \frac{7}{10}F_{A}\\ 10y=7.35-7y\\ y=0.432m$

The Load should be suspended at a point 43.2 cm from the wire A such that there is an equal strain in the two wires.

Let the ends of the steel wire be called A and B.

length of the wire is 2l = 1 m.

The cross-sectional area of the wire is $A=0.50\times 10^{-2}cm^{2}$

Let the depression at the midpoint due to the suspended 100 g be y.

Change in the length of the wire is $\Delta l$

$\\\Delta l=AD+DB-AB\\ \Delta l=\sqrt{l^{2}+y^{2}}+\sqrt{l^{2}+y^{2}}-2l\\ \Delta l=2\sqrt{l^{2}+yx^{2}}-2l\\ \Delta l=2l\left (\left ( 1+\frac{y^{2}}{l^{2}} \right )^{\frac{1}{2}}-1 \right )\\ \Delta l=2l(1+\frac{y^{2}}{2l^{2}}-1)\\ \Delta l=\frac{y^{2}}{l}$

The strain is $\frac{\Delta l}{2l}=\frac{y^{2}}{2l^{2}}$

The vertical components of the tension in the arms balance the weight of the suspended mass, we have

$\\2Fcos\theta =mg\\ 2F\frac{y}{\sqrt{l^{2}+y^{2}}}=mg\\ F=\frac{mg\sqrt{l^{2}+y^{2}}}{2y}\\ F=\frac{mgl}{2y}$

The stress in the wire will be

$\frac{F}{A}=\frac{mgl}{2Ay}$

The Young's Modulus of steel is $Y=2\times 10^{11}Nm^{-2}$

$\\Y=\frac{Stress}{Strain}\\ Y=\frac{\frac{mgl}{2Ay}}{\frac{y^{2}}{2l^{2}}}\\ Y=\frac{mgl^{3}}{Ay^{3}}\\$

$\\y=\left ( \frac{mgl^{3}}{AY} \right )^{\frac{1}{3}}\\ y=\left ( \frac{100\times 10^{-3}\times 9.8\times (0.5)^{3}}{0.5\times 10^{-2}\times 10^{-4}\times 2\times 10^{11}} \right )\\ y=1.074\times 10^{-2}m\\ y=1.074cm$

The depression at the mid-point of the steel wire will be 1.074 cm.

Diameter of each rivet, d = 6.0 mm

Maximum Stress $P=6.9\times 10^{7}Nm^{-2}$

The number of rivets, n = 4.

The maximum tension that can be exerted is T

$\\T=nP\pi \left ( \frac{d}{2} \right )^{2}\\ T=4\times 6.9\times 10^{7}\times \pi \times \left ( \frac{6\times 10^{-3}}{2} \right )^{2}\\ T=7803.71N$

The pressure at the bottom of the trench, $P=1.1\times 10^{8}Pa$

The initial volume of the steel ball, V = 0.32 m

Bulk Modulus of steel, $B=1.6\times 10^{11}Nm^{-2}$

$\\B=\frac{P}{\frac{\Delta V}{V}}\\ \Delta V=\frac{PV}{B}\\ \Delta V=\frac{1.1\times 10^{8}\times 0.32}{1.6\times 10^{11}}\\ \Delta V=2.2\times 10^{-4}m^{3}$

The change in the volume of the ball when it reaches the bottom of the trench is  $2.2\times 10^{-4}m^{3}$.

## NCERT solutions for class 11 physics chapter wise:

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## NCERT Solutions for Class 11 Subject wise:

 CBSE NCERT solutions for class 11 biology NCERT solutions for class 11 maths CBSE NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Importance of NCERT solutions for class 11 physics chapter 9 mechanical properties of solids:

By practicing NCERT solutions for class 11 physics chapter 9 mechanical properties of solids, you will be able to answer most of the numerical problems asked in final exams for class 11 and other competitive exams like NEET and JEE mains. The chapter mechanical properties of solids comes under the unit properties of bulk matter for NEET and properties of solids and liquids for JEE Mains. Two or three questions are expected for NEET from the unit properties of bulk matter out of which one question is expected from the chapter mechanical properties of solids and in JEE Mains one question may be asked from mechanical properties of solids.