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

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

Edited By Vishal kumar | Updated on Sep 05, 2023 09:00 AM IST

NCERT Solutions Class 11 Physics Chapter 9 – Free PDF Download

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Solids is an invaluable resource that aids students in comprehending the topics and preparing effectively for their exams. class 11 Physics Chapter 9 exercise solutions chapter consists of a total of twenty-one questions, each of which is explained in great detail and in a straightforward manner in Class 11 Physics Ch 9 NCERT solutions. It is crucial to take comprehensive notes while studying each topic. Solving textbook questions and gaining a strong grasp of the subject matter are equally vital. The experienced faculty has meticulously crafted the NCERT solutions in alignment with the latest CBSE Syllabus for 2023, with the aim of enhancing students' conceptual understanding.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

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  1. NCERT Solutions Class 11 Physics Chapter 9 – Free PDF Download
  2. NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids
  3. NCERT Solutions for Class 11 Physics Chapter Wise
  4. Importance of Mechanical Properties Of Solids Class 11 Solutions
  5. Key features of mechanical properties of solids exercise solutions
  6. Subject wise NCERT Exemplar solutions

Class 11 Physics Chapter 9 NCERT solutions are part of NCERT Solutions for Class 11 Physics. If a rubber band is stretched by applying a variable force after a particular amount of force the rubber band breaks. 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 Mechanical Properties Of Solids Class 11 solutions mainly deal 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. Class 11 Physics Chapter 9 question answer helps students in their preparations. NCERT is considered the bible for many competitive exams.

Mechanical properties of solids class 11 exercise pdf download for free.

**According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 9, " Mechanical properties of solids," has been renumbered as Chapter 8.

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

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Access NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids Exercise

Q1 A steel wire of length 4.7 m and cross-sectional area 3.0 \times 10^{-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 10^{-5} m^2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

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 = 3.0 \times 10^{-5} m^2

The cross-sectional area of the Copper wire A C = 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.

Q2 (a) Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus 1650444678674

Answer:

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}

Q2 (b) Figure 9.11 shows the strain-stress curve for a given material. What are approximate yield strength for this material?

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Answer:

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

Q3 (a) The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale. Which of the materials has the greater Young’s modulus?

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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.

Q3 (b) The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale. Which of the two is the stronger material?
Fig. 9.12

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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.

Q4 (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.

Q4 (b) Read the following two statements below carefully and state, with reasons, if it is true or false. The stretching of a coil is determined by its shear modulus.

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.

Q5 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.

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Answer:

Tension in the steel wire is F 1

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

Length of steel wire l 1 = 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, Y 1 = 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 F 2

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

Length of Brass wire l 2 = 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, Y 2 = 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

Q6 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?

Answer:

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

Area of a face of the Aluminium cube, A = l 2 = 0.01 m 2

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

Q7 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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer:

Inner radii of each column, r 1 = 30 cm = 0.3 m

Outer radii of each colum, r 2 = 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}

Q8 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?

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, \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}

Q9 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?

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

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

Q10 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 ratios of their diameters if each is to have the same tension.

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.

\\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

Q11 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 cross-sectional 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.

Answer:

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 cm 2

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

Q12 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 10 ^ 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.

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

\\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.

Q13 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 103 Kg ?

Answer:

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} .

Q14 Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm.

Answer:

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}

Q15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 \times 10 ^ 6 Pa

Answer:

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 = s 3

V = (0.1) 3

V = 0.001 m 3

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.

Q16 How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answer:

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?

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Answer:

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

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Answer:

Cross-Sectional Area of wire A is A A = 1 mm 2

Cross-Sectional Area of wire B is A B = 2 mm 2

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

Let the Tension in the two wires A and B be F A and F B 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.

Q18 (b) 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 mm^2 \: \:and \: \: \: 2.0 mm^2
respectively. At what point along the rod should a mass m be suspended in order to produce equal strains in both steel and aluminium wires.

1650445144040

Answer:

Cross-Sectional Area of wire A is A A = 1 mm 2

Cross-Sectional Area of wire B is A B = 2 mm 2

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

Let the Tension in the two wires A and B be F A and F B 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.

Q19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 \times 10 ^{-2} cm^2 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.

Answer:

1650445189592

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.

Q20 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.

Answer:

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

Q21 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 0.32 m^3 is 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 to the bottom?

Answer:

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 3

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} .

Class 11 Physics Chapter 9 exercise solutions hold significant importance for students. Questions related to the Mechanical Properties of Solids are commonly featured in Class 11 examinations, making it a crucial topic for those aiming to achieve high marks in both school exams and entrance exams. To enhance their understanding of this chapter, students are encouraged to solve the mechanical properties of solids exercise provided at the end of the chapter.

Mechanical properties of solids class 11 exercise explore the fundamental concept of determining a solid's mechanical properties, a central theme in Physics. Students will delve into topics such as Young's modulus for materials like copper and steel. They will learn to extract valuable information, including yield strength, from graph plots and compare stress-strain relation graphs to determine Young's modulus and strength values.

By mastering the content in this class 11 physics ch 9 ncert solutions, students will not only excel in their academic examinations but also gain a deeper insight into the mechanical behaviour of materials, a skill that can be valuable in various real-world applications.

NCERT Solutions for Class 11 Physics Chapter Wise

Chapter 1

Physical world

Chapter 2

Units and Measurement

Chapter 3

Motion in a straight line

Chapter 4

Motion in a Plane

Chapter 5

Laws of Motion

Chapter 6

Work, Energy and Power

Chapter 7

System of Particles and Rotational motion

Chapter 8

Gravitation

Chapter 9

Mechanical Properties of Solids

Chapter 10

Mechanical Properties of Fluids

Chapter 11

Thermal Properties of Matter

Chapter 12

Thermodynamics

Chapter 13

Kinetic Theory

Chapter 14

Oscillations

Chapter 15

Waves

Mechanical properties of solids numericals: Important Formulas and Diagrams + eBook link

Here are some important formulas and diagrams for mechanical properties of solids class 11 exercise. You can also access the eBook link for comprehensive study materials.

  • Stress=\frac{Force}{Area}

  • Hooke's law E=\frac{Stress }{Strain}

    E= modulus \: of \: elasticity

  • longitudinal strain

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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}

r = Radius of wire

L = Original length

\Delta l = Change in length

  • Pressure

Pressure applied by a liquid column, p = hρg

  • Viscous Force

(F)= -η A (dv/dx)

here,

(dv/dx) = rate of change of the velocity with distance and velocity gradient, A = area of cross-section and coefficient of viscosity

  • Variation in Viscosity

nt= no/(1+αt+βt2)

where no and nt are coefficients of viscosities at 0°C and t°C, α and β are constants

  • Poiseuille’s Formula

ν= π/2(pr4/ηl)

Where:

p = pressure difference across the two ends of the tube, r = radius of the tube,n = coefficient of viscosity and l = length of the tube

These are some of the important formulas that play a crucial role in understanding the concepts covered in NCERT Solutions class 11 physics chapter 9. Utilize them effectively for your studies and problem-solving.

Importance of Mechanical Properties Of Solids Class 11 Solutions

By practising NCERT solutions for class 11, 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 come 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.

Key features of mechanical properties of solids exercise solutions

Comprehensive Coverage: NCERT class 11 physics chapter 9 question answer cover all the topics and concepts present in the chapter, ensuring a thorough understanding of mechanical properties of solids.

Step-by-Step Solutions: Detailed step-by-step numericals on mechanical properties of solids class 11 for each question are provided, making it easy for students to follow and grasp the concepts.

Clarity and Simplicity: The mechanical properties of solids numericals are explained in a clear and simple language, making complex concepts easy to understand.

Diagrams and Graphs: Wherever necessary, diagrams and graphs are included to enhance visual understanding and clarify concepts.

Concise Summaries: NCERT mechanical properties of solids exercise solutions often include concise summaries of the main points in each section, helping students revise and retain important information.

Examination-Oriented: The mechanical properties of solids class 11 numericals are designed to help students prepare for exams, including CBSE board exams and various competitive exams.

NCERT Solutions for Class 11 Subject wise

NCERT solutions for Class 11 Biology

NCERT solutions for Class 11 Maths

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What are the topics covered in the Class 11 chapter Mechanical Properties of Solids

The topics covered in the NCERT book Class 11 physics chapter 9 are

  • Elastic behaviour of solids
  • Stress and strain
  • Hooke’s law
  • Stress-strain curve
  • Elastic moduli
  • Applications of elastic behaviour of materials
2. What is the weightage of the chapter mechanical properties of solids for JEE Main exam

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.

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0.34\; J

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0.16\; J
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1.00\; J

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2.45×10−3 kg

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0.02

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less than 3

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