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

Ans : \(\begin{array}{l}{\text { Length of the steel wire, } L_{1}=4.7 \mathrm{m}} \\ {\text { Area of cross-section of the steel wire, } A_{1}=3.0 \times 10^{-5} \mathrm{m}^{2}} \\ {\text { Length of the copper wire, } L_{2}=3.5 \mathrm{m}} \\ {\text { Area of cross-section of the copper wire, } \mathrm{A}_{2}=4.0 \times 10^{-5} \mathrm{m}^{2}}\end{array}\)
\(\begin{array}{l}{\text { Length of the copper wire, } L_{2}=3.5 \mathrm{m}} \\ {\text { Area of cross-section of the copper wire, } \mathrm{A}_{2}=4.0 \times 10^{-5} \mathrm{m}^{2}} \\ {\text { Change in length }=\Delta L_{1}=\Delta L_{2}=\Delta L} \\ {\text { Force applied in both the cases }=F} \\ {\text { Young's modulus of the steel wire: }}\end{array}\)
\(\begin{array}{l}{Y_{1}=\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L}} \\ {=\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L}_{\ldots(i)}}\end{array}\)
\(\begin{array}{l}{\text { Young's modulus of the copper wire: }} \\ {Y_{2}=\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}} \\ {=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}}\end{array}\)
\(\begin{array}{l}{\text { Dividing (i) by (ii), we get: }} \\ {\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}=1.79 : 1} \\ {\text { The ratio of Young's modulus of steel to that of copper is } 1.79 : 1 .}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (a) It is clear from the given graph that for stress } 150 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}, \text { strain is } 0.002} \\ {=\frac{\text { Stress }}{\text { Strain }}} \\ {\therefore \text { Young's modulus, } Y} \\ {=\frac{150 \times 10^{6}}{0.002}=7.5 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}}\end{array}\)
\(\begin{array}{l}{\text { Hence, Young's modulus for the given material is } 7.5 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} \text { . }} \\ {\text { (b) The yield strength of a material is the maximum stress that the material can sustain }} \\ {\text { without crossing the elastic limit. }} \\ {\text { It is clear from the given graph that the approximate yield strength of this material is }} \\ {300 \times 10^{6} \mathrm{Nm} /^{2} \text { or } 3 \times 10^{8} \mathrm{N} / \mathrm{m}^{2}}\end{array}\) 

Ans : (a) For a given strain, the stress for material A is more than it is for material B, as 
shown in the two graphs. 
Young's modulus \(\begin{array}{l}{=\frac{\text { Stress }}{\text { Strain }}} \\ {\text { Young's modulus }} \\ {\text { For a given strain, if the stress for a material is more, then Young's modulus is also }} \\ {\text { greater for that material. Therefore, Young's modulus for material A is greater than it is }} \\ {\text { for material B. }}\end{array}\)
\(\begin{array}{l}{\text { (b) The amount of stress required for fracturing a material, corresponding to its fracture }} \\ {\text { point, gives the strength of that material. Fracture point is the extreme point in a stress- }} \\ {\text { strain curve. It can be observed that material A can withstand more strain than material }} \\ {\text { B. Hence, material A is stronger than material B. }}\end{array}\) 

Ans : (a) False (b) True 
(a) For a given stress, the strain in rubber is more than it is in steel. 
Young's modulus, \(Y=\frac{\text { Stress }}{\text { Strain }}\)
For a constant stress: \(Y \propto \frac{1}{\text { Strain }}\)
Hence, Young's modulus for rubber is less than it is for steel. 
(b) Shear modu us is the ratio of the applied stress to the change in the shape of a 
body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is 
involved in this process. 

Ans : \(\begin{array}{l}{\text { Elongation of the steel wire }=1.49 \times 10^{-4} \mathrm{m}} \\ {\text { Elongation of the brass wire }=1.3 \times 10^{-4} \mathrm{m}} \\ {\text { Diameter of the wires, } d=0.25 \mathrm{m}}\end{array}\)
\(\begin{array}{l}{\text { Hence, the radius of the wires, }^{r=\frac{d}{2}}=0.125 \mathrm{cm}} \\ {\text { Length of the steel wire, } L_{1}=1.5 \mathrm{m}} \\ {\text { Length of the brass wire, } L_{2}=1.0 \mathrm{m}} \\ {\text { Total force exerted on the steel wire: }} \\ {F_{1}=(4+6) \mathrm{g}=10 \times 9.8=98 \mathrm{N}} \\ {\text { Young's modulus for steel: }}\end{array}\)
\(Y_{1}=\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{\Delta L_{1}}{L_{1}}\right)}\)
\(\begin{array}{l}{\text { Where, }} \\ {\Delta L_{1}=\text { Change in the length of the steel wire }} \\ {A_{1}=\text { Area of cross-section of the steel wire }=\pi r_{1}^{2}} \\ {\text { Young's modulus of steel, } Y_{1}=2.0 \times 10^{11} \text { Pa }}\end{array}\)
\(\begin{array}{l}{\therefore \Delta L_{1}=\frac{F_{1} \times L_{1}}{A_{1} \times Y_{1}}=\frac{F_{1} \times L_{1}}{\pi r_{1}^{2} \times Y_{1}}} \\ {=\frac{98 \times 1.5}{\pi\left(0.125 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}}=1.49 \times 10^{-4} \mathrm{m}}\end{array}\)
\(\begin{array}{l}{\text { Total force on the brass wire: }} \\ {F_{2}=6 \times 9.8=58.8 \mathrm{N}} \\ {\text { Young's modulus for brass: }} \\ {Y_{2}=\frac{\left(\frac{F_{2}}{A_{2}}\right)}{\left(\frac{\Delta L_{2}}{L_{2}}\right)}}\end{array}\)
\(\begin{array}{l}{\text { Where, }} \\ {\Delta L_{2}=\text { Change in length }} \\ {A_{2}=\text { Area of cross-section of the brass wire }}\end{array}\)
\(\begin{array}{l}{\therefore \Delta L_{2}=\frac{F_{2} \times L_{2}}{A_{2} \times Y_{2}}=\frac{F_{2} \times L_{2}}{\pi r_{2}^{2} \times Y_{2}}} \\ {=\frac{58.8 \times 1.0}{\pi \times\left(0.125 \times 10^{-2}\right)^{2} \times\left(0.91 \times 10^{11}\right)}=1.3 \times 10^{-4} \mathrm{m}}\end{array}\)
\(\begin{array}{l}{\text { Elongation of the steel wire }=1.49 \times 10^{-4} \mathrm{m}} \\ {\text { Elongation of the brass wire }=1.3 \times 10^{-4} \mathrm{m}}\end{array}\)