Chapter 8: Mechanical Properties of Solids

Ans: In the above question it is given that:

The length of the steel wire is ${{L}_{1}}=4.7m$.

Area of cross-section of the steel wire is \[{{A}_{1}}=3.0\times {{10}^{-5}}{{m}^{2}}\].

The length of the copper wire is ${{L}_{2}}=3.5m$.

Area of cross-section of the copper wire is \[{{A}_{2}}=4.0\times {{10}^{-5}}{{m}^{2}}\] .

Now,

The change in length is given by:

$\Delta L={{L}_{1}}-{{L}_{2}}=4.7-3.5=1.2m$

Let the force applied in both the cases be $F$ .

Therefore, Young’s modulus of the steel wire is given by:

${{Y}_{1}}=\frac{F}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta L}=\frac{F\times 4.7}{3.0\times {{10}^{-5}}\times 1.2}$ …… (1)

And Young’s modulus of the copper wire is given by:

${{Y}_{2}}=\frac{F}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta L}=\frac{F\times 3.5}{4.0\times {{10}^{-5}}\times 1.2}$ …… (2)

Therefore,

$\frac{{{Y}_{1}}}{{{Y}_{2}}}=\frac{F\times 4.7\times 4.0\times {{10}^{-5}}\times 1.2}{3.0\times {{10}^{-5}}\times 1.2\times F\times 3.5}=\frac{1.79}{1}$

Hence the ratio of Young’s modulus of steel to that of copper is $1.79:1$ .

Ans: From the graph given in the above question it is clear that:

Stress for a given material is $150\times {{10}^{6}}N/{{m}^{2}}$ and strain is $0.002$.

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{150\times {{10}^{6}}N/{{m}^{2}}}{0.002}=7.5\times {{10}^{10}}N/{{m}^{2}}$

Therefore, Young’s modulus for the given material is $7.5\times {{10}^{10}}N/{{m}^{2}}$ .

Ans: The yield strength of the material is the maximum stress it sustains without crossing the elastic limit.

From the graph given in the above question, it is clear that approximate yield strength for this material is $300\times {{10}^{6}}N/{{m}^{2}}$ or $3\times {{10}^{8}}N/{{m}^{2}}$ .

Ans:  In the two graphs it is that given that stress for A is more than that of B.

As, 

\[Young's\text{ }modulus=\frac{Stress}{Strain}\],

Therefore, material A has greater Young's modulus.

Ans: The strength of a material is determined by the amount of stress required for fracturing a material, corresponding to its fracture point. 

Fracture point is defined as the extreme point in a stress-strain curve. 

From the graph it is clear that material A can withstand more strain than material B. 

Therefore, material A is stronger than material B.

Ans: The given statement is false. 

As there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain. Therefore, the Young’s modulus of steel is greater than that of rubber.

Ans: The given statement is true. 

As the shear modulus of a coil relates with the change in shape of the coil and the stretching of coil changes its shape without any change in the length. Therefore, the shear modulus of elasticity is involved. Hence the stretching of a coil is determined by its shear modulus.

Ans: In the above question it is given that:

Diameter of the wires is $d=0.25m$.

Hence $r=0.125m$.

Length of the steel wire is ${{L}_{1}}=1.5m$.

Length of the brass wire is ${{L}_{2}}=1.0m$.

Total force exerted on the steel wire is ${{F}_{1}}=\left( 4+6 \right)g=10g$

$\therefore {{F}_{1}}=10\times 9.8=98N$.

Young’s modulus for steel is given by 

${{Y}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}\times \frac{{{L}_{1}}}{\Delta {{L}_{1}}}$

Where,

$\Delta {{L}_{1}}$ is the change in the length of the steel wire.

And ${{A}_{1}}$ is the area of cross-section of the steel wire.

$\therefore {{A}_{1}}=\pi {{r}_{1}}^{2}$

We have,

Young’s modulus of steel is ${{Y}_{1}}=2.0\times {{10}^{11}}Pa$.

$\Rightarrow \Delta {{L}_{1}}=\frac{{{F}_{1}}\times {{L}_{1}}}{{{A}_{1}}\times {{Y}_{1}}}$

$\Rightarrow \Delta {{L}_{1}}=\frac{98\times 1.5}{\pi {{\left( 0.125 \right)}^{2}}\times 2.0\times {{10}^{11}}}=1.49\times {{10}^{-4}}m$.

Total force on the brass wire is ${{F}_{2}}=6\times 9.8=58.8N$.

Young’s modulus for brass is given by 

${{Y}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}\times \frac{{{L}_{2}}}{\Delta {{L}_{2}}}$

Where,

$\Delta {{L}_{2}}$ is the change in the length of the brass wire.

And ${{A}_{2}}$ is the area of cross-section of the brass wire.

$\therefore {{A}_{2}}=\pi {{r}_{2}}^{2}$

We have,

Young’s modulus of brass is ${{Y}_{2}}=0.91\times {{10}^{11}}Pa$.

$\Rightarrow \Delta {{L}_{2}}=\frac{{{F}_{2}}\times {{L}_{2}}}{{{A}_{2}}\times {{Y}_{2}}}$

$\Rightarrow \Delta {{L}_{2}}=\frac{58.8\times 1}{\pi {{\left( 0.125 \right)}^{2}}\times 0.91\times {{10}^{11}}}=1.3\times {{10}^{-4}}m$.

Clearly, the elongation of the steel wire is $1.49\times {{10}^{-4}}m$ and that of the brass wire is $1.3\times {{10}^{-4}}m$.

Ans: In the above question it is given that:

Edge of the aluminium cube is $L=10cm=0.1m$.

The mass attached to the cube is $m=100kg$.

Shear modulus $\eta $ of aluminium is $25GPa=25\times {{10}^{10}}Pa$.

We know that:

\[Shear\text{ }modulus\left( \eta  \right)=\frac{\text{Shear stress}}{\text{Shear strain}}\]

$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{\left( \frac{L}{\Delta L} \right)}$

Where,

$F$ is the applied force.

$\therefore F=mg=100\times 9.8=980N$

Area of one of the faces of the cube is $A=0.1\times 0.1=0.01{{m}^{2}}$.

Vertical deflection of the cube is $\Delta L$.

$\Delta L=\frac{FL}{A\eta }$

$\Rightarrow \Delta L=\frac{980\times 0.1}{0.01\times 25\times {{10}^{9}}}=3.92\times {{10}^{-7}}m$

Clearly, the vertical deflection of this face of the cube is $3.92\times {{10}^{-7}}m$.

Ans: In the above question it is given that:

Mass of the big structure is $M=50000kg$.

Inner radius of the column is $r=30cm=0.3m$.

Outer radius of the column is $R=60cm=0.6m$ 

Young’s modulus of steel is $Y=2\times {{10}^{11}}Pa$

The total force exerted is $F=Mg=50000\times 9.8N$.

\[\text{Stress = Force exerted on a single column}\]

$\Rightarrow Stress=\frac{50000\times 9.8}{4}=122500N$

Young’s modulus is given by:

 $Y=\frac{Stress}{Strain}$

$\Rightarrow Strain=\frac{\left( \frac{F}{A} \right)}{Y}$

Where,

Area is given by

$A=\pi \left( {{R}^{2}}-{{r}^{2}} \right)=\pi \left( {{\left( 0.6 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right)$.

$\Rightarrow Strain=\frac{\left( \frac{50000\times 9.8}{\pi \left( {{\left( 0.6 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right)} \right)}{2\times {{10}^{11}}}=7.22\times {{10}^{-7}}$

Therefore, the compressional strain of each column is $7.22\times {{10}^{-7}}$.

Ans: In the above question it is given that:

Length of the piece of copper is $l=19.1mm=19.1\times {{10}^{-3}}m$.

Breadth of the piece of copper is $b=15.2mm=15.2\times {{10}^{-3}}m$

Area of the copper piece will be:

$A=l\times b$

$\Rightarrow A=19.1\times {{10}^{-3}}\times 15.2\times {{10}^{-3}}=2.9\times {{10}^{-4}}{{m}^{2}}$

Tension force applied on the piece of copper is $F=44500N$.

Modulus of elasticity of copper is $\eta =42\times {{10}^{9}}N/{{m}^{2}}$.

We know that :

\[\text{Modulus of elasticity}\left( \eta  \right)=\frac{Stress}{Strain}\]

$\Rightarrow \eta =\frac{\left( \frac{F}{A} \right)}{Strain}$

$\Rightarrow Strain=\frac{F}{A\eta }$

$\Rightarrow Strain=\frac{44500}{2.9\times {{10}^{-4}}\times 42\times {{10}^{9}}}=3.65\times {{10}^{-3}}$

Hence, the resulting strain is $3.65\times {{10}^{-3}}$.

Ans: In the above question it is given that:

Radius of the steel cable is $r=1.5cm=0.015m$.

Maximum allowable stress is ${{10}^{8}}N/{{m}^{2}}$.

We know that:

\[\text{Maximum force = Maximum stress }\times \text{Area of cross-section}\]

\[\Rightarrow \text{Maximum force =1}{{\text{0}}^{8}}\times \pi {{\left( 0.015 \right)}^{2}}=7.065\times {{10}^{4}}N\]

Therefore, the cable can support the maximum load of \[7.065\times {{10}^{4}}N\].

Ans: In the above question it is given that:

Tension force acting on each wire is the same.

Therefore the extension produced in each wire is the same.

As the length of both wires is the same, the strain in both wires is also the same.

Young’s modulus is given by:

$Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{Strain}=\frac{\frac{4F}{\pi {{d}^{2}}}}{Strain}$ …… (1)

Where,

$F$ is the Tension force,

$A$ is the area of cross-section and

$d$ is the diameter of the wire

From equation (1), it is clear that $Y\propto \frac{1}{{{d}^{2}}}$.

Young’s modulus for iron is ${{Y}_{1}}=190\times {{10}^{9}}Pa$.

Let the diameter of the iron wire be ${{d}_{1}}$.

Young’s modulus for copper is ${{Y}_{2}}=100\times {{10}^{9}}Pa$.

Let the diameter of the copper wire be ${{d}_{2}}$.

Therefore, the ratio of their diameters is given as:

$\frac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\frac{{{Y}_{1}}}{{{Y}_{2}}}}=\sqrt{\frac{190\times {{10}^{9}}}{100\times {{10}^{9}}}}=\frac{1.31}{1}$

Therefore, the ratio of diameters of copper wire to iron wire is $1.31:1$.