# Mechanical Properties of Solids

## Class 11 NCERT Physics

### NCERT

1   A steel wire of length $4.7 m$ and cross-sectional area $3.0 * 10^{-5} m ^2$ stretches by the same amount as a copper wire of length $3.5 m$ and cross-sectional area of $4.0 * 10^{-5} m^ 2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Length of the steel wire, $L_1 = 4.7 m$$\\ Area of cross-section of the steel wire,A_1=3.0*10^{-5}m^2$$\\$ Length of the copper wire, $L_2 = 3.5 m$$\\ Area of cross-section of the copper wire, A_2 = 4.0 * 10^{-5} m^2$$\\$ Change in length =$\Delta L_1 =\Delta L _2= \Delta L$$\\ Force applied in both the cases = F$$\\$ Young’s modulus of the steel wire:$\\$ $Y_1=\dfrac{F}{A_1}*\dfrac{L_1}{\Delta L}\\ =\dfrac{F*4.7}{3.0*10^{-5}*\Delta L}....(i)$$\\ Young’s modulus of the copper wire:\\ Y_2=\dfrac{F}{A_2}*\dfrac{L_2}{\Delta L}\\ =\dfrac{F*4.7}{4.0* 10^{-5} * \Delta L}....(ii)$$\\$ Dividing (i) by (ii), we get:$\\$ $\dfrac{Y_1}{Y_2}=\dfrac{4.7*4*10^{-5}}{3*10^{-5}*3.5}\\ =1.79$$\\ The ratio of Young’s modulus of steel to that of copper is 1.79 : 1. 2 Figure shows the strain-stress curve for a given material. What are \\ (a) Young’s modulus and\\ (b) approximate yield strength for this material? ##### Solution : (a) It is clear from the given graph that for stress 150 *10^6 N / m^2 , strain is 0.002.$$\\$ $\therefore$ Young’s modulus, Y = Stress / Strain $\\$ $=150 *10^6 / 0.002 = 7.5 * 10^{10} Nm^{-2}$$\\ Hence, Young’s modulus for the given material is 7.5 * 10^{10} N / m^2 .\\ (b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.\\ It is clear from the given graph that the approximate yield strength of this material is\\ 300*10^6 N / m^2 or 3*10^8 N / m^2 . 3 The stress-strain graphs for materials A and B are shown in figure. The graphs are drawn to the same scale.\\ (a) Which of the materials has the greater Young’s modulus?\\ (b) Which of the two is the stronger material? ##### Solution : (a) From the two graphs, we note that for a given strain, stress for A is more than that of B. Hence, Young's modulus (\dfrac{\text{stress}}{\text{strain}}) is greater for A than that of B.\\ (b) The amount of stress required for fracturing a material, corresponding to its fracture point, gives the strength of that material. Fracture point is the extreme point in a stress-strain curve. It can be observed that material A can withstand more strain than material B. Hence, material A is stronger than material B. 4 Read the following two statements below carefully and state, with reasons, if it is true or false.\\ (a) The Young’s modulus of rubber is greater than that of steel;\\ (b) The stretching of a coil is determined by its shear modulus. ##### Solution : (a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.\\ (b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved. 5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. ##### Solution : Elongation of the steel wire = 1.49 *10 -4 m$$\\$ Elongation of the brass wire $=1.3 * 10 -4 m$$\\ Diameter of the wires, d = 0.25 m$$\\$ Hence, the radius of the wires, $r = d/2 = 0.125 cm$$\\ Length of the steel wire, L_1 = 1.5 m$$\\$ Length of the brass wire, $L_2 = 1.0 m$$\\ Total force exerted on the steel wire:\\ F_1=(4 + 6) g = 10 * 9.8 = 98 N$$\\$ Young’s modulus for steel:$\\$ $Y_1 (F_1 / A_1 ) / (\Delta L_1 / L_1 )$$\\ Where,\\ L_1 = Change in the length of the steel wire\\ A_1 = Area of cross-section of the steel wire =\pi r_1^2$$\\$ Young’s modulus of steel, $Y_1 = 2.0 * 10^{11} Pa $$\\ \therefore L_1 = F_1 * L _1 / ( A_1 * Y_1 )\\ =(98 * 1 .5 ) / [ \pi( 0.125 * 10^{-2})* 2 * 10^{11} ] \\ = 1.49 * 10^{-4} m$$\\$ Total force on the brass wire:$\\$ $F_2 = 6 *9.8 =58.8 N$$\\ Young’s modulus for brass:\\ Y_2 =( F_2 / A_2) / (\Delta _ 2 / L_2 )$$\\$ Where,$\\$ $\Delta L_2 =$ Change in the length of the brass wire$\\$ $A_1 =$ Area of cross-section of the brass wire $=\pi r_1 ^2$$\\ \therefore L_2 = F_2 * L_2 / (A_2 * Y_2 )\\ =(58.8 * 1 )/ [ ( \pi *(0.125 * 10^{-2} )*( 0.91 * 10^{11} ) ] \\ = 1.3 *10^{- 4} m$$\\$ Elongation of the steel wire $= 1.49 * 10^{-4} m$$\\ Elongation of the brass wire = 1.3 *10^{-4} m .$$\\$

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?

Edge of the aluminium cube,$L = 10 cm = 0.1 m$$\\ The mass attached to the cube, m = 100 kg$$\\$ Shear modulus $(\eta)$ of aluminium $=25 GPa = 25 * 10 ^9 Pa$$\\ Shear modulus, \eta = Shear stress / Shear strain =(F / A ) / ( L / \Delta L )$$\\$ Where,$\\$ F = Applied force = mg = $100 × 9.8 = 980 N$$\\ A = Area of one of the faces of the cube=0.1 * 0.1 = 0.01 m^2$$\\$ $\Delta L$= Vertical deflection of the cube$\\$ $\therefore \Delta L=FL/A\eta\\ =980*0.1/[10^{-2}*(25*10^9)]\\ =3.92*10^{-7}m$$\\ The vertical deflection of this face of the cube is 3.92*10^{-7}m 7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. ##### Solution : Mass of the big structure, M = 50,000 kg$$\\$ Inner radius of the column,$r = 30 cm = 0.3 m$$\\ Outer radius of the column, R = 60 cm = 0.6 m$$\\$ Young’s modulus of steel, $Y = 2 * 10 ^{11} Pa$$\\ Total force exerted, F = Mg = 50000 × 9.8 N$$\\$ Stress = Force exerted on a single column $= 50000 × 9.8 / 4 = 122500 N$$\\ Young’s modulus, Y = Stress / Strain\\ Where,\\ Area,A=\pi(R^2-r^2)=\pi ((0.6)^2-(0.3)^2)$$\\$ Strain=$122500/[\pi ((0.6)^2-(0.3)^2)*2*10^{11}]=7.22*10^{-7}$$\\ Hence,the compressional strain of each column is 7.22*10^{-1}$$\\$

8   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?