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.
Hooke's law
longitudinal strain
The ratio of change in length to the original length.
r = Radius of wire
L = Original length
= Change in length
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} =
The cross-sectional area of the Copper wire A_{C} =
Let the load and the change in the length be F and respectively
Since F and is the same for both wires we have
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
Young’s modulus is given as the ratio of stress to strain when the body is behaving elastically.
For the given material
The Yield Strength is approxmiately 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.
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
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;
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 F_{1}
Length of steel wire l_{1} = 1.5 m
The diameter of the steel wire, d = 0.25 cm
Area od the steel wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y_{1} =
Tension in the Brass wire is F_{2}
Length of Brass wire l_{2} = 1.5 m
Area od the brass wire,
Let the elongation in the steel wire be
Young's Modulus of steel, Y_{2} =
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
Tangential Stress is F/A
Shear modulus of aluminium
Let the Vertical deflection be
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
Youngs Modulus of steel is
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,
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
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.
As F, l and are equal for all wires
Mass of the body = 14.5 kg
Angular velocity, = 2 rev/s
The radius of the circle, r = 1.0 m
Tension in the wire when the body is at the lowest point is T
Cross-Sectional Area of wire, A = 0.065 cm^{2}
Young's Modulus of steel,
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
The bulk modulus of air is
The Ratio of the Bulk Modulus of water to that of air is
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
Bulk Modulus of water is
The negative sign signifies that for the same given mass the Volume has decreased
The density of water at the surface
Let the density at the given depth be
Let a certain mass occupy V volume at the surface
Dividing the numerator and denominator of RHS by V we get
The density of water at a depth where pressure is 80.0 atm is .
Bulk's Modulus of Glass is
Pressure is P = 10 atm.
The fractional change in Volume would be given as
The fractional change in Volume is
The bulk modulus of copper is
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
From the definition of bulk modulus
The volumetric strain is
Volume contraction will be
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%?
Change in volume is
Bulk modulus of water is
A pressure of 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.
The diameter of at the end of the anvil, d = 0.50 mm
Cross-sectional area at the end of the anvil is A
Compressional Force applied, F = 50000N
The pressure at the tip of the anvil is P
The pressure at the tip of the anvil is .
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
Equating moments of the Tension in the wires about the point where mass m is suspended we have
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 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
Equating moments of the Tension in the wires about the point where mass m is suspended we have
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
Let the depression at the midpoint due to the suspended 100 g be y.
Change in the length of the wire is
The strain is
The vertical components of the tension in the arms balance the weight of the suspended mass, we have
The stress in the wire will be
The Young's Modulus of steel is
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
The number of rivets, n = 4.
The maximum tension that can be exerted is T
The pressure at the bottom of the trench,
The initial volume of the steel ball, V = 0.32 m^{3 }
Bulk Modulus of steel,
The change in the volume of the ball when it reaches the bottom of the trench is .
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.
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 A steel ball of initial volume 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?
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?
Q16 How much should the pressure on a litre of water be changed to compress it by 0.10%?