## NCERT Solutions Class 11 Physics Chapter 9 - Mechanical Properties of SolidsThe NCERT Answers Class 11 Physics Chapter 9 Mechanical Properties of Solids are crucial tools for students to learn the material and get ready for the test. Making notes while carefully studying each topic is crucial. It's crucial to complete the textbook exercises and learn the material thoroughly. With the intention of enhancing conceptual knowledge among students, the knowledgeable staff created the NCERT Answers in accordance with the most recent CBSE Syllabus 2022-23. Although though it's common knowledge that a rigid body has a set size and shape, it can actually be squeezed, stretched, and bent. A force is required to alter the form of a solid. The two primary ideas in this chapter that need attention are elasticity and plasticity. ## NCERT Solutions Class 11 Physics Chapter 9
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10
For steel wire: L For steel wire: L Thus,the force applied F and the extension (ΔL) will be same for both the wires. Young's modulus of steel can be written as,
The figure below shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
a.) Young's modulus is given by the equation, Y = Stress Strain The stress and strain within the limit of proportionality is 150 × 10 Substituting the values in the above equation we get, b.) The greatest stress value at which there is permanent deformation is known as the material's yield strength. The maximum value of yield strength from the given graph is approximately 3× 10
The stress-strain graphs for materials A and B are shown in the figure below. The graphs are drawn to the same scale. (a) Which of the materials has the greater Young's modulus? (b) Which of the two is the stronger material?
a.) Young's modulus is given by the equation, Y= stress strain At a particular strain, Young's modulus is directly proportional to the stress. The stress strain curve for material A has greater stress at particular strain. Therefore, the material A has greater Young's modulus than the material B. b.) The material's strength is influenced by the stress at the rupture or fracture point. More than material B, material A has a higher fracture strength. Thus, substance A is more powerful.
Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young's modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.
a.) The statement is false. The equation, which yields Young's modulus, Y = stress strain Young's modulus has an inverse relationship with strain for a given stress. When rubber is under a certain tension, its strain is greater than that of steel. Rubber is less elastic than steel because it has a lower Young's modulus. b.) The assertion is true. Shear tension develops in the coil as it is compressed or stretched. The material's shear modulus is, G stands for shear stress and strain. As a result, the coil's ability to stretch is dependent upon the material's shear modulus of stiffness.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass, are loaded, as shown in Fig. 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. [Young's modulus of steel is 2.0 x 10
We need to find the total force exerted on the steel wire and brass wire. Total force exerted on the steel wire,
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?
Four identical hollow cylindrical columns of mild steel support a big structure with a mass of 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.
Mass of the big structure,M = 50,000 kg Total force exerted on the four columns, = Total weight of the structure = 50000×9.8N The compressional force on each column,
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.
Given the dimension of piece of copper are 15.22 mm × 19.1 mm. Applied force, = 44500 N The Modulu of elasticity of copper is = 140 × 10
A steel cable with a radius of 1.5 cm supports a chairlift in a ski area. If the maximum stress is not to exceed 10
The radius of the steel cable is = 1.5 cm Maximum stress = 10 Maximum stress is given by,
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.
Each wire is subject to the exact same tensile force. As a result, the extension is the same in each instance. The strain will also be the same because the wires are the same length. The relation of Young's modulus is given by, Young's modulus of iron, Y Diameter of the iron wire=d Young's modulus of copper, Y Diameter of the copper wire=d Thus, ration of their diameters is given as,
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
Mass of the object,m = 14.5 Kg Length of the steel wire,l = 1 m Angular velocity, ω = 2 rev/s = 2 × 2π = 12.56 rad/s Cross-sectional area of the wire,A = 0.065 cm When a mass is spun in a vertical circle, the net force imposed on it at its lowest point is the total of the forces exerted by gravity and centrifugal force.
Compute the bulk modulus of water from the following data: Initial volume = 100.0 litres, Pressure increase = 100.0 atm (1 atm = 1.013 × 10
Initial volume of the water,V Final volume of the water,V Increase in atmospheric pressure, ΔP = 100.0 atm = 100 × 1.013 × 10 Change in water volume, ΔV = V Bulk modulus is given by the equation, Since liquid molecules are closer together than those in air, the intermolecular force in liquids is significantly greater than that in air. As a result, at the same temperature, water is under significantly higher strain than air.
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 × 10
Pressure of water at a depth = 80 atm Density of water at the surface = 1.03 × 10 Let V Let the density of water at the suface be = ρ Let V Let the density of water at the depth be = ρ
Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm.
Hydraulic pressure exerted on the glass slab P = 10 atm = 10 × 1.013 × 10 Bulk modulus of glass, B = 37 × 10
Determine the volume contraction of a solid copper cube, 10 cm on edge, when subjected to a hydraulic pressure of 7.0 × 10
Length of an edge of the solid copper cube,l = 10 cm = 0.1 m Hydraulic pressure, p = 7.0 × 10 Buluk modulus of copper, B = 140 × 10
How much should the pressure on a litre of water be changed to compress it by 0.10%?
Volume of water, V = 1 L Water needs to be compressed by 0.01%
Anvils made of single crystals of diamond, with the shape as shown in the figure, are used to investigate the 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?
Diameter of the anvil = 0.50 mm Compressive force acting on the anvil = 50000 N Presure at the tip of the anvil, P = FA = 4Fπd Substituting the values inthe above equation, P = 4 × 50000 × π × (0.5× 10 Thus, the pressure at the tip of the anvil is 2.5 × 10
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 the figure. The cross-sectional areas of wires A and B are 1.0 mm
Cross - sectional area of wire A, a Cross - sectional area of wire B, b We know the Young's modulus of steel, Y Young's modulus of steel, Y (i) Let a mass m be hanged at a distance y from the end of the stick where wire A is fastened. It is given that both the wires have equal stresses. Where, F F The above equation can be presented using the following picture. Moment of force about the point of suspension, we have In order to produce an equal amount of strain, let the mass m be suspended from the stick at a distance y Considering the force at the location where mass m is suspended. In order to cause equal strain in the two wires, the mass must be suspended at a distance of 0.432 m from the end where wire A is attached.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10
From the above figure, Let x be the depression at the mid point i.e.CD = x. From figure Ac = CB = 1 = 0.5 m mass, m = 100 g = 0.100 Kg
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 × 10
Diameter of each rivet is = 6.0 mm Shear stress on the rivet is = 6.9 × 10 The equation, gives the shear stress on a rivet. = maxmum stress × cross - sectional area = 6.9 × 10 Maximum tension = 4 × 1950 N = 7800 N
The Marina trench is located in the Pacific Ocean, and in one place, it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 × 10
Bulk-modulus of steel,(B) = 160 × 10 Initial volume of steel ball,V = 0.32 m Water pressure at the bottom, p = 1.1 × 10 At a depth of 11 km beneath the surface of the Pacific Ocean, the ball crashes to the bottom. When the ball reaches the bottom of the trench,let its volume change be ΔV. The ball's volume changes by 2.148 × 10
A mild steel wire of cross-sectional area 0.60 x 10
Let YZ be the mild steel wire of length 2l = 2m Cross-sectional area A= 0.60 × 10 Let the mass of m = 100 g = 0.1 Kg be hung from the midpoint of O,as shown in the figure. Let x be the depression at the midpoint,i.e.,OD |