Answer

Length of the steel wire, $L_1=4.7 m$

Area of cross-section of the steel wire, $A_1=3.0 \times 10^{-5} m^{2}$

Length of the copper wire, $L_2=3.5 m$

Area of cross-section of the copper wire, $A_2=4.0 \times 10^{-5} m^{2}$

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:

$$ \begin{align*} & Y_1=\frac{F_1}{A_1} \times \frac{L_1}{\Delta L} \\ & =\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \tag{i} \end{align*} $$

Young’s modulus of the copper wire:

$$ \begin{align*} Y_2 & =\frac{F_2}{A_2} \times \frac{L_2}{\Delta L_2} \\ & =\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L} \tag{ii} \end{align*} $$

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 $

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

Answer

It is clear from the given graph that for stress $150 \times 10^{6} N / m^{2}$, strain is 0.002 .

$\therefore$ Young’s modulus, $Y=\frac{\text{ Stress }}{\text{ Strain }}$

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

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

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 $\times 10^{6} Nm /{ }^{2}$ or $3 \times 10^{8} N / m^{2}$.

Answer

(a) A

(b) A

For a given strain, the stress for material $\mathbf{A}$ is more than it is for material $\mathbf{B}$, as shown in the two graphs.

Young’s modulus $=\frac{\text{ Stress }}{\text{ Strain }}$

For a given strain, if the stress for a material is more, then Young’s modulus is also greater for that material. Therefore, Young’s modulus for material A is greater than it is for material $\mathbf{B}$.

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 stressstrain curve. It can be observed that material $\mathbf{A}$ can withstand more strain than material $\mathbf{B}$. Hence, material $\mathbf{A}$ is stronger than material $\mathbf{B}$.

Answer

(a) False

(b) True

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.

Shear modulus 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.

Answer

Elongation of the steel wire $=1.49 \times 10^{-4} m$

Elongation of the brass wire $=1.3 \times 10^{-4} m$

Diameter of the wires, $d=0.25 m$

Hence, the radius of the wires, $\quad r=\frac{d}{2}=0.125 cm$

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:

$F_1=(4+6) g=10 \times 9.8=98 N$

Young’s modulus for steel:

$Y_1=\frac{(\frac{F_1}{A_1})}{(\frac{\Delta L_1}{L_1})}$

Where,

$\Delta L_1=$ 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, $Y_1=2.0 \times 10^{11} Pa$

$ \begin{aligned} \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(0.125 \times 10^{-2})^{2} \times 2 \times 10^{11}}=1.49 \times 10^{-4} m \end{aligned} $

Total force on the brass wire:

$F_2=6 \times 9.8=58.8 N$

Young’s modulus for brass:

$Y_2=\frac{(\frac{F_2}{A_2})}{(\frac{\Delta L_2}{L_2})}$

Where,

$\Delta L_2=$ Change in length $A_2=$ Area of cross-section of the brass wire

$\therefore \Delta L_2=\frac{F_2 \times L_2}{A_2 \times Y_2}=\frac{F_2 \times L_2}{\pi r_2^{2} \times Y_2}$

$=\frac{58.8 \times 1.0}{\pi \times(0.125 \times 10^{-2})^{2} \times(0.91 \times 10^{11})}=1.3 \times 10^{-4} m$

Elongation of the steel wire $=1.49 \times 10^{-4} m$

Elongation of the brass wire $=1.3 \times 10^{-4} m$

Answer

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 $=25 GPa=25 \times 10^{9} Pa$

$ =\frac{\text{ Shear stress }}{\text{ Shear strain }}=\frac{\frac{F}{A}}{L} $

Shear modulus, $\eta$

Where,

$F=$ Applied force $=m g=100 \times 9.8=980 N$

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

$\Delta L=$ Vertical deflection of the cube

$\therefore \Delta L=\frac{F L}{A \eta}$

$ =\frac{980 \times 0.1}{10^{-2} \times(25 \times 10^{9})} $

$=3.92 \times 10^{-7} m$

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

Answer

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, $Y=2 \times 10^{11} Pa$

Total force exerted, $F=M g=50000 \times 9.8 N$

Stress $=$ Force exerted on a single column $=\frac{50000 \times 9.8}{4}=122500 N$

Young’s modulus, $Y=\frac{\text{ Strcss }}{\text{ Strain }}$

Strain $=\frac{\frac{F}{A}}{Y}$

Where,

Area, $A=\pi(R^{2}-r^{2})=\pi((0.6)^{2}-(0.3)^{2})$

Strain $=\frac{122500}{\pi[(0.6)^{2}-(0.3)^{2}] \times 2 \times 10^{11}}=7.22 \times 10^{-7}$

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

Answer

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

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

Area of the copper piece:

$A=l \times b$

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

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

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

Modulus of elasticity, $\eta=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\frac{F}{A}}{\text{ Strain }}$

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

$ =\frac{44500}{2.9 \times 10^{-4} \times 42 \times 10^{9}} $

$=3.65 \times 10^{-3}$

Answer

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

Maximum allowable stress $=10^{8} N m^{-2}$

Maximum stress $=\frac{\text{ Maximum force }}{\text{ Area of cross-section }}$

$\therefore$ Maximum force $=$ Maximum stress $\times$ Area of cross-section

$=10^{8} \times \pi(0.015)^{2}$

$=7.065 \times 10^{4} N$

Hence, the cable can support the maximum load of $7.065 \times 10^{4} N$.

Answer

The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.

The relation for Young’s modulus is given as:

$$ \begin{equation*} Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\frac{F}{A}}{\text{ Strain }}=\frac{\frac{4 F}{\pi d^{2}}}{\text{ Strain }} \tag{i} \end{equation*} $$

Where,

$F=$ Tension force

$A=$ Area of cross-section

$d=$ Diameter of the wire

It can be inferred from equation $(i)$ that $Y \propto \frac{1}{d^{2}}$

Young’s modulus for iron, $Y_1=190 \times 10^{9} Pa$

Diameter of the iron wire $=d_1$

Young’s modulus for copper, $Y_2=110 \times 10^{9} Pa$

Diameter of the copper wire $=d_2$

Therefore, the ratio of their diameters is given as:

$\frac{d_2}{d_1}=\sqrt{\frac{Y_1}{Y_2}}=\sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}}=\sqrt{\frac{19}{11}}=1.31: 1$

Answer

Mass, $m=14.5 kg$

Length of the steel wire, $l=1.0 m$

Angular velocity, $\omega=2 rev / s$

Cross-sectional area of the wire, $a=0.065 cm^{2}$

Let $\delta l$ be the elongation of the wire when the mass is at the lowest point of its path.

When the mass is placed at the position of the vertical circle, the total force on the mass is:

$ \begin{aligned} & F=m g+m l \omega^{2} \\ & =14.5 \times 9.8+14.5 \times 1 \times(2)^{2} \\ & =200.1 N \\ & \text{ Young’s modulus }=\frac{\text{ Stress }}{\text{ Strain }} \\ & Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{F}{A} \frac{l}{\Delta l} \\ & \therefore \Delta l=\frac{F l}{A Y} \end{aligned} $

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

$ \begin{aligned} \therefore \Delta l & =\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}}=1539.23 \times 10^{-7} \\ & =1.539 \times 10^{-4} m \end{aligned} $

Hence, the elongation of the wire is $1.539 \times 10^{-4} m$.

Answer

Initial volume, $V_1=100.01=100.0 \times 10^{-3} m^{3}$

Final volume, $V_2=100.51=100.5 \times 10^{-3} m^{3}$

Increase in volume, $\Delta V=V_2-V_1=0.5 \times 10^{-3} m^{3}$

Increase in pressure, $\Delta p=100.0 atm=100 \times 1.013 \times 10^{5} Pa$

Bulk modulus $=\frac{\Delta p}{\frac{\Delta V}{V_1}}=\frac{\Delta p \times V_1}{\Delta V}$

$ \begin{aligned} & =\frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{-3}}{0.5 \times 10^{-3}} \\ & =2.026 \times 10^{9} Pa \end{aligned} $

Bulk modulus of air $=1.0 \times 10^{5} Pa$

$\therefore \frac{\text{ Bulk modulus of water }}{\text{ Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}}=2.026 \times 10^{4}$

This ratio is very high because air is more compressible than water.

Answer

Let the given depth be $h$.

Pressure at the given depth, $p=80.0 atm=80 \times 1.01 \times 10^{5} Pa$

Density of water at the surface, $\rho_1=1.03 \times 10^{3} kg m^{-3}$

Let $\rho_2$ be the density of water at the depth $h$.

Let $V_1$ be the volume of water of mass $m$ at the surface.

Let $V_2$ be the volume of water of mass $m$ at the depth $h$.

Let $\Delta V$ be the change in volume.

$ \begin{aligned} \Delta V & =V_1-V_2 \\ & =m(\frac{1}{\rho_1}-\frac{1}{\rho_2}) \end{aligned} $

$\therefore$ Volumetric strain $=\frac{\Delta V}{V_1}$

$ =m(\frac{1}{\rho_1}-\frac{1}{\rho_2}) \times \frac{\rho_1}{m} $

$\therefore \frac{\Delta V}{V_1}=1-\frac{\rho_1}{\rho_2}$

Bulk modulus, $B=\frac{p V_1}{\Delta V}$

$ \frac{\Delta V}{V_1}=\frac{p}{B} $

Compressibity of water $=\frac{1}{B}=45.8 \times 10^{-11} Pa^{-1}$

$$ \begin{equation*} \therefore \frac{\Delta V}{V_1}=80 \times 1.013 \times 10^{5} \times 45.8 \times 10^{-11}=3.71 \times 10^{-3} \tag{ii} \end{equation*} $$

For equations ( $i$ ) and (ii), we get:

$ \begin{aligned} & 1-\frac{\rho_1}{\rho_2}=3.71 \times 10^{-3} \\ & \rho_2=\frac{1.03 \times 10^{3}}{1-(3.71 \times 10^{-3})} \\ & \quad=1.034 \times 10^{3} kg m^{-3} \end{aligned} $

Therefore, the density of water at the given depth $(h)$ is $1.034 \times 10^{3} kg m^{-3}$.

Answer

Hydraulic pressure exerted on the glass slab, $p=10 atm=10 \times 1.013 \times 10^{5} Pa$

Bulk modulus of glass, $B=37 \times 10^{9} Nm^{-2}$

Bulk modulus, $B=\frac{p}{\Delta V}$

Where,

$ \begin{aligned} & \frac{\Delta V}{V}=\text{ Fractional change in volume } \\ & \begin{aligned} \therefore \frac{\Delta V}{V} & =\frac{p}{B} \\ & =\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}} \\ & =2.73 \times 10^{-5} \end{aligned} \end{aligned} $

Hence, the fractional change in the volume of the glass slab is $2.73 \times 10^{-5}$.

Answer

Length of an edge of the solid copper cube, $l=10 cm=0.1 m$

Hydraulic pressure, $p=7.0 \times 10^{6} Pa$

Bulk modulus of copper, $B=140 \times 10^{9} Pa$

Bulk modulus, $B=\frac{p}{\frac{\Delta V}{V}}$

Where,

$\frac{\Delta V}{V}=$ Volumetric strain

$\Delta V=$ Change in volume

$V=$ Original volume.

$\Delta V=\frac{p V}{B}$

Original volume of the cube, $V=l^{3}$

$\therefore \Delta V=\frac{p l^{3}}{B}$

$ \begin{aligned} & =\frac{7 \times 10^{6} \times(0.1)^{3}}{140 \times 10^{9}} \\ & =5 \times 10^{-8} m^{3} \\ & =5 \times 10^{-2} cm^{-3} \end{aligned} $

Therefore, the volume contraction of the solid copper cube is $5 \times 10^{-2} cm^{-3}$.

Answer

Volume of water, $V=1 L$

It is given that water is to be compressed by $0.10 \%$. $\therefore$ Fractional change, $\frac{\Delta V}{V}=\frac{0.1}{100 \times 1}=10^{-3}$

Bulk modulus, $B=\frac{\rho}{\Delta V}$

$p=B \times \frac{\Delta V}{V}$

Bulk modulus of water, $B=2.2 \times 10^{9} Nm^{-2}$

$ \begin{aligned} p & =2.2 \times 10^{9} \times 10^{-3} \\ & =2.2 \times 10^{6} Nm^{-2} \end{aligned} $

Therefore, the pressure on water should be $2.2 \times 10^{6} Nm^{-2}$.