Sample Exercises

Hint

For steel
\(
l_1=4.7 \mathrm{~m}, \quad A_1=3.0 \times 10^{-5} \mathrm{~m}^2
\)
If \(F\) newton is the stretching force and \(\Delta l\) metre the extension in each case, then
\(
\begin{aligned}
Y_1 & =-\frac{F l_1}{A_1 \Delta l} \dots(i)\\
\Rightarrow \quad Y_1 & =\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta l}
\end{aligned}
\)
For copper
Now,
\(
l_2=3.5 \mathrm{~m}, \quad A_2=4.0 \times 10^{-5} \mathrm{~m}^2
\)
\(Y_2=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta l} \dots(ii)\)
Dividing (i) by (ii), we get
\(
\frac{Y_1}{Y_2}=\frac{4.7}{3.0 \times 10^{-5}} \times \frac{4.0 \times 10^{-5}}{3.5}=\frac{4.7 \times 4.0}{3.0 \times 3.5}=1.79 :1
\)

Hint

(a) Young’s modulus of the material \((Y)\) is given by
\(
\begin{aligned}
& Y=\text { Stress } / \text { Strain } \\
& =150 \times 10^6 / 0.002 \\
& 150 \times 10^6 / 2 \times 10-3 \\
& =75 \times 10^9 \mathrm{Nm}^{-2} \\
& =75 \times 10^{10} \mathrm{Nm}^{-2}
\end{aligned}
\)

(b)Yield strength of a material is defined as the maximum stress it can sustain. From graph, the approximate yield strength of the given material
\(
\begin{aligned}
& =300 \times 10^6 \mathrm{Nm}^{-2} \\
& =3 \times 10^8 \mathrm{Nm}^{-2} .
\end{aligned}
\)

Hint

(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 \(=(\) Stress \(/\) Strain \()\) is greater for \(A\) than that of \(B\).

(b) Strength of a material is determined by the amount of stress required to cause a fracture. This stress corresponds to the point of fracture. The stress corresponding to the point of fracture in A is more than for B. So, material A is stronger than material B.

Hint

(a) False. The-Young’s modulus is defined as the ratio of stress to the strain within the elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True.  The stretching of the coil simply changes its shape without any change in the length of the wire used and shear modulus deals with the change in the shape of the material.

Hint

For steel wire; total force on steel wire;
\(
\begin{aligned}
F_1 & =4+6=10 \mathrm{~kg} f=10 \times 9.8 \mathrm{~N} ; \\
l_1 & =1.5 \mathrm{~m}, \quad \Delta l_1=? ; \quad 2 r_1=0.25 \mathrm{cm}
\end{aligned}
\)
or

\(
r_1=\left(\frac{0.25}{2}\right) \mathrm{cm}=0.125 \times 10^{-2} \mathrm{~m}
\)
\(
\begin{aligned}
Y_1=2.0 \times 10^{11} \mathrm{~Pa}
\end{aligned}
\)
For brass wire,
\(
\begin{aligned}
F_2 & =6.0 \mathrm{~kg} f=6 \times 9.8 \mathrm{~N} ; \\
2 r_2 & =0.25 \mathrm{~cm} \\
r_2 & =\left(\frac{0.25}{2}\right)=0.125 \times 10^{-2} \mathrm{~m} ; \\
Y_2 & =0.91 \times 10^{11} \mathrm{~Pa}, \quad l_2=1.0 \mathrm{~m}, \Delta l_2=?
\end{aligned}
\)
Since,
or
\(
Y_1=\frac{F_1 \times l_1}{A_1 \times \Delta l_1}=\frac{F_1 \times l_1}{\pi r_1^2 \times \Delta l_1} \Rightarrow \Delta l_1=\frac{F_1 \times l_1}{\pi r_1^2 \times \Delta l_1}
\)
or
\(
\Delta l_1=\frac{(10 \times 9.8) \times 1.5 \times 7}{22 \times\left(0.125 \times 10^{-2}\right)^2 \times 2 \times 10^{11}}=1.49 \times 10^{-4} \mathrm{~m} .
\)
and
\(\Delta l_2=\frac{F_2 \times l_2}{\pi r_2^2 \times Y_2}=\frac{(6 \times 9.8) \times 1 \times 7}{22 \times\left(0.125 \times 10^{-2}\right)^2 \times\left(0.91 \times 10^{11}\right)}=1.3 \times 10^{-4} \mathrm{~m}\).

Hint

\(
\begin{aligned}
& G=\frac{F \times L}{A \times \Delta l} \\
& \therefore \Delta l=\frac{m \times g \times L}{A \times G} \\
& \therefore \Delta l=\frac{100 \times 9.8 \times 0.1}{0.1 \times 0.1 \times 25 \times 10^{9}} \\
& \therefore \Delta l=3.92 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)

Hint

Here total mass to be supported, \(M=50,000 \mathrm{~kg}\)
\(\therefore\) Total weight of the structure to be supported \(=\mathrm{Mg}\)
\(
=50,000 \times 9.8 \mathrm{~N}
\)
Since this weight is to be st ported by 4 columns,
\(\therefore\) Compressional force on each column \((F)\) is given by
\(
F=\frac{\mathrm{Mg}}{4}=\frac{50,000 \times 9.8}{4} \mathrm{~N}
\)
Inner radius of a column, \(r_1=30 \mathrm{~cm}=0.3 \mathrm{~m}\)
Outer radius of a column, \(r_2=60 \mathrm{~cm}=0.6 \mathrm{~m}\).
\(\therefore\) Area of cross-section of each column is given by
\(
\begin{aligned}
A & =\pi\left(r_2^2-r_1^2\right) \\
& =\pi\left[(0.6)^2-(0.3)^2\right]=0.27 \pi \mathrm{m}^2
\end{aligned}
\)
Young’s modulus, \(Y=2 \times 10^{11} \mathrm{~Pa}\)
Compressional strain of each column \(=\) ?
\(
\begin{aligned}
\therefore \quad Y & =\frac{\text { Compressional force } / \text { area }}{\text { Compressional Strain }} \\
& =\frac{F / A}{\text { Compressional Strain }}
\end{aligned}
\)
or Compressional strain of each column
\(
\begin{aligned}
& =\frac{F}{A Y}=\frac{50,000 \times 9.8 \times 7}{4 \times 0.27 \times 22 \times 2 \times 10^{11}} \\
& =0.722 \times 10^{-6}
\end{aligned}
\)
\(\therefore\) Compressional strain of all columns is given by
\(
\begin{aligned}
& =0.722 \times 10^{-6} \times 4=2.88 \times 10^{-6} \\
& =2.88 \times 10^{-6} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A=15.2 \times 19.2 \mathrm{~mm}^2=15.2 \times 19.2 \times 10^{-6} \mathrm{~m}^2 \\
& \mathrm{~F}=44,500 \mathrm{~N} \\
& \text { Strain = F/AY }=44500 / 15.2 \times 19.1 \times 10^{-6} \times 1.2 \times 10^{11} \\
& =0.001277
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { Maximum load } & =\text { Maximum stress } \times \text { Cross-sectional area } \\
& =10^8 \mathrm{Nm}^{-2} \times \frac{22}{7} \times\left(1.5 \times 10^{-2} \mathrm{~m}\right)^2 \\
& =7.07 \times 10^4 \mathrm{~N} .
\end{aligned}
\)

Hint

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:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}=\frac{\frac{4 F}{\pi d^2}}{\text { Strain }} \text {….(i) }
\)
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, \(Y_1=190 \times 10^9 \mathrm{~Pa}\)
Diameter of the iron wire \(=d_1\)
Young’s modulus for copper, \(Y_2=120 \times 10^9 \mathrm{~Pa}\)
Diameter of the copper wire \(=d_2\)
Therefore, the ratio of their diameters is given as:
\(
\frac{d_2}{d_1}=\sqrt{\frac{Y_1}{Y_2}}=\sqrt{\frac{190 \times 10^9}{120 \times 10^9}}=\sqrt{\frac{19}{12}}=1.25
\)

Hint

Mass, \(\mathrm{m}=14.5 \mathrm{~kg}\)
Length of the steel wire, \(\mathrm{l}=1.0 \mathrm{~m}\)
Angular velocity, \(\omega=2\) rev \(/ \mathrm{s}\)
Cross-sectional area of the wire, \(a=0.065 \mathrm{~cm}^2\)
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:
\(
\begin{aligned}
& F=m g+m l \omega^2 \\
& =14.5 \times 9.8+14.5 \times 1 \times(2)^2=200.1 \mathrm{~N}
\end{aligned}
\)
\(
\text { Young’s modulus }=\frac{\text { Stress }}{\text { Strain }}
\)
\(
Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{F}{A} \frac{l}{\Delta l}
\)
\(
\therefore \Delta l=\frac{F l}{A Y}
\)
Young’s modulus for steel \(=2 \times 10^{11} \mathrm{~Pa}\)
\(
\begin{aligned}
\therefore \Delta l & =\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}}=1539.23 \times 10^{-7} \\
& =1.539 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)
Hence, the elongation of the wire is \(1.539 \times 10^{-4} \mathrm{~m}\).

Hint

Initial volume, \(\mathrm{V}_1=100.01=100.0 \times 10^{-3} \mathrm{~m}^3\)
Final volume, \(\mathrm{V}_2=100.5 \mathrm{l}=100.5 \times 10^{-3} \mathrm{~m}^3\)
Increase in volume, \(\Delta \mathrm{V}=\mathrm{V}_2-\mathrm{V}_1=0.5 \times 10^{-3} \mathrm{~m}^3\)
Increase in pressure, \(\Delta \mathrm{p}=100.0 \mathrm{~atm}=100 \times 1.013 \times 10^5 \mathrm{~Pa}\)
\(
\begin{aligned}
\text { Bulk modulus } & =\frac{\Delta p}{\frac{\Delta V}{V_1}}=\frac{\Delta p \times V_1}{\Delta V} \\
& =\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}} \\
& =2.026 \times 10^9 \mathrm{~Pa}
\end{aligned}
\)
Bulk modulus of air \(=1.0 \times 10^5 \mathrm{~Pa}\)
\(
\therefore \frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^9}{1.0 \times 10^5}=2.026 \times 10^4
\)
This ratio is very high because air is more compressible than water.

Hint

Let the given depth be \(h\).
Pressure at the given depth, \(p=80.0 \mathrm{~atm}=80 \times 1.01 \times 10^5 \mathrm{~Pa}\)
Density of water at the surface, \(\rho_1=1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
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.
\(
\begin{aligned}
& \Delta V=V_1-V_2 \\
& =m\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right) \\
& \therefore \text { Volumetric strain }=\frac{\Delta V}{V_1}
\end{aligned}
\)
\(
\begin{aligned}
& =m\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right) \times \frac{\rho_1}{m} \\
& \therefore \frac{\Delta V}{V_1}=1-\frac{\rho_1}{\rho_2} \ldots \text {…(i) } \\
& \text { Bulk modulus, } B=\frac{p V_1}{\Delta V} \\
& \frac{\Delta V}{V_1}=\frac{p}{B} \\
& \text { Compressibity of water }=\frac{1}{B}=45.8 \times 10^{-11} \mathrm{~Pa}^{-1} \\
& \therefore \frac{\Delta V}{V_1}=80 \times 1.013 \times 10^5 \times 45.8 \times 10^{11}=3.71 \times 10^{-3} \dots(ii)
\end{aligned}
\)
For equation (i) and (ii) we get
\(
\begin{aligned}
& 1-\frac{\rho_1}{\rho_2}=3.71 \times 10^{-3} \\
& \rho_2=\frac{1.03 \times 10^3}{1-\left(3.71 \times 10^{-3}\right)} \\
& =1.034 \times 10^3 \mathrm{kgm}^{-3}
\end{aligned}
\)
Therefore, the density of water at the given depth \((h)\) is \(1.034 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\).

Hint

Hydraulic pressure exerted on the glass slab, \(p=10 \mathrm{~atm}=10 \times 1.013 \times 10^5 \mathrm{~Pa}\)
Bulk modulus of glass, \(B=37 \times 10^9 \mathrm{Nm}^{-2}\)
Bulk modulus, \(B=\frac{\mathrm{p}}{\Delta \mathrm{V} / \mathrm{v}}\)
where,
\(\Delta V / V=\) Fractional change in volume
\(
\begin{aligned}
\therefore \Delta V / V & =p / B \\
& =\frac{10 \times 1.013 \times 10^5}{\left(37 \times 10^9\right)} \\
& =2.73 \times 10^{-5}
\end{aligned}
\)
Hence, the fractional change in the volume of the glass slab is \(2.73 \times 10^{-5}\).

Hint

\(
\begin{aligned}
& \mathrm{P}=7.0 \times 10^6 \mathrm{~Pa} \\
& \mathrm{l}=10 \mathrm{~cm} \\
& \mathrm{~V}=\mathrm{l}^3=10^{-3} \mathrm{~m}^3 \\
& \mathrm{~B}=140 \times 10^9 \\
& \mathrm{~B}=\frac{\mathrm{P} \times \mathrm{V}}{\Delta \mathrm{V}} \\
& \Delta \mathrm{V}=\frac{\mathrm{P}}{\mathrm{B}} \times \mathrm{V}=\frac{7 \times 10^6}{140 \times 10^9} \times 10^{-3} \mathrm{~m}^3 \\
& =5 \times 10^{-8} \mathrm{~m}^3 \\
& =0.05 \mathrm{~cm}^3
\end{aligned}
\)

Hint

\(
\mathrm{B}=\frac{\mathrm{P} \times \mathrm{V}}{\Delta \mathrm{V}}
\)

\(
p=B \frac{\Delta V}{V}=\left(2.2 \times 10^9\right) \times 10^{-3}=2.2 \times 10^6 \mathrm{~Pa}
\)

Hint

Diameter of the corner end of the anvil,
\(
d=0.50 \mathrm{~mm}=0.50 \times 10^{-3} \mathrm{~m}
\)
Area of cross-section of tip,
\(
\begin{aligned}
A & =\frac{\pi d^2}{4} \\
& =\frac{22 \times\left(0.50 \times 10^{-3}\right)^2}{7 \times 4} \mathrm{~m}^2
\end{aligned}
\)
Stress (= pressure at the tip of the anvil)
\(
\begin{aligned}
& =\frac{F}{A}=\frac{50,000 \times 4 \times 7}{22 \times(0.50)^2 \times 10^{-6}} \mathrm{Nm}^{-2} \\
& =2.54 \times 10^{11} \mathrm{Nm}^{-2}(\text { or } \mathrm{Pa}) .
\end{aligned}
\)

Hint

For steel wire \(A, l_1=l_{;} ; A_1=1 \mathrm{~mm}^2 ; Y_1=2 \times 1011 \mathrm{Nm}^{-2}\)
For aluminium wire \(B, I_2=I ; A_2=2 \mathrm{~mm}^2 ; Y_2=7 \times 1010 \mathrm{Nm}^{-2}\)
a) Let mass \(m\) be suspended from the rod at distance \(x\) from the end where wire \(A\) is connected. Let \(F_1\) and \(\mathrm{F}_2\) be the tensions in two wires and there is equal stress in two wires, then
\(
\frac{F_1}{A_1}=\frac{F_2}{A_2} \Rightarrow \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1}{2} \ldots \text {.. (i) }
\)
Taking moment of forces about the point of suspension of mass from the rod, we have
\(
F_1 x=F_2(1.05-x) \text { or } \frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2}
\)
or \(2.10-2 x=x=>x=0.70 \mathrm{~m}=70 \mathrm{~cm}\)

(b) Let mass \(m\) be suspended from the rod at distance \(x\) from the end where wire \(A\) is connected. Let \(F_1\) and \(\mathrm{F}_2\) be the tension in the wires and there is equal strain in the two wires i.e.,
\(
\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2} \Rightarrow \frac{F_1}{F_2}=\frac{A_1 Y_1}{A_2 Y_2}=\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}
\)
As the rod is stationary, so \(F_1 x=F_2(1.05-x)\) or
\(
\begin{aligned}
& \frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{10}{7} \\
& =>10 \mathrm{x}=7.35-7 \mathrm{x} \text { or } \mathrm{x}=0.4324 \mathrm{~m}=43.2 \mathrm{~cm}
\end{aligned}
\)

Hint

Let \(A B\) be a mild steel wire of length \(2 L=I \mathrm{~m}\) and its cross-section area \(A=0.50 \times 10^{-2} \mathrm{~cm}^2\). A mass \(\mathrm{m}=100\) \(\mathrm{g}=0.1 \mathrm{~kg}\) is suspended at mid-point \(\mathrm{C}\) of wire as shown in figure. Let \(\mathrm{x}\) be the depression at mid-point i.e., \(C D=x\)
\(
\therefore A D=D B=\sqrt{\left(A C^2+C D^2\right)}=\sqrt{L^2+x^2}
\)
\(\therefore\) Increase in length \(\triangle L=(A D+D B)-A B=2 \sqrt{L^2+x^2}-2 L\)
\(
=2 L\left[\left(1+\frac{x^2}{L^2}\right)^{\frac{1}{2}}-1\right]=2 L \cdot \frac{x^2}{2 L^2}=\frac{x^2}{L}
\)
\(\therefore\) Longitudinal strain \(=\frac{\Delta L}{2 L}=\frac{x^2}{2 L^2}\)
If \(\mathrm{T}\) be the tension in the wire as shown figure then in equilibrium \(2 T \cos \theta=m g\)
or \(T=\frac{\mathrm{mg}}{2 \cos \theta}\)
\(
\begin{aligned}
& =\frac{\mathrm{mg}}{2 \frac{x}{\sqrt{x^2+L^2}}}=\frac{m g \sqrt{x^2+L^2}}{2 x}=\frac{\mathrm{mgL}}{2 x} \\
& \therefore \text { Stress }=\frac{T}{A}=\frac{\mathrm{mgL}}{2 x A}
\end{aligned}
\)
\(
\begin{aligned}
& \text { As Young’s modulus } Y=\frac{\text { stress }}{\text { strain }} \\
& =\frac{\frac{m g L}{2 x A}}{\frac{x^2}{2 L^2}}=\frac{m g L}{2 x A} \times \frac{2 L^2}{x^2}=\frac{m g L^3}{A x^3} \\
& \Rightarrow x=\left[\frac{m g L^3}{Y A}\right]^{\frac{1}{3}}=L\left[\frac{m g}{Y A}\right]^{\frac{1}{3}} \\
& =\frac{1}{2}\left[\frac{0.1 \times 9.8}{2 \times 10^{11} \times 0.50 \times 10^{-2} \times 10^{-4}}\right]^{\frac{1}{3}}=1.074 \times 10^{-2} \mathrm{~m} \\
& =1.074 \mathrm{~cm} \approx 1.07 \mathrm{~cm} \text { or } 0.01 \mathrm{~m}
\end{aligned}
\)

Hint

Let the tension exerted by riveted strip \(=\mathrm{F}\)
This tension would provide shearing force on the four rivets, which share it equally.
\(\therefore\) Shearing force on each rivet \(=F / 4\)
and shearing stress on each rivet \(=F / 4 / A=F / 4 A\)
As the maximum shearing stress on each rivet is given to be \(2.3 \times 10^9 \mathrm{~Pa}\), so we have
\(
\begin{aligned}
& \frac{F_{\text {max }}}{4 A}=2.3 \times 10^9 \\
& \text { or } F_{\max }=4 A \times 2.3 \times 10^9 \\
& =4 \times \pi r^2 \times 2.3 \times 10^9 \\
& =4 \times \frac{22}{7} \times\left(3.0 \times 10^3\right)^2 \times 2.3 \times 10^9 \\
& =260.2 \times 10^3 \mathrm{~N}=260 \mathrm{kN}
\end{aligned}
\)

Hint

Water pressure at the bottom of the trench is given as
\(
\mathrm{p}=1.1 \times 10^8 \mathrm{~Pa}
\)
Initial volume of the steel ball is given as \(V=0.32 \mathrm{~m}^3\)
Bulk modulus of steel is known as \(\mathrm{B}=1.6 \times 10^{11} \mathrm{Nm}^{-2}\)
The ball lands at the bottom of the trench, which is about 11 kilometres below the water’s surface.
Let \(\triangle \mathrm{V}\) be the volume change of the ball after reaching the bottom of the trench.
Bulk modulus, \(B=p /(\Delta V / V)=p V / \Delta V\)
\(
\begin{aligned}
& \Delta V=p V / B \\
& =\left(1.1 \times 10^8 \times 0.32\right) /\left(1.6 \times 10^{11}\right) \\
& =2.2 \times 10^{-4} \mathrm{~m}^3
\end{aligned}
\)

Hint

We have
\(
Y=\frac{\text { stress }}{\text { strain }}=\frac{T / A}{l / L}
\)
with symbols having their usual meanings. The extension is
\(
l=\frac{T L}{A Y} .
\)
As the load is in equilibrium after the extension, the tension in the wire is equal to the weight of the load
\(
=4.8 \mathrm{~kg} \times 10 \mathrm{~m} \mathrm{~s}^{-2}=48 \mathrm{~N}
\)
Thus, \(l=\frac{(48 \mathrm{~N})(2 \mathrm{~m})}{\left(0.2 \times 10^{-4} \mathrm{~m}^2\right) \times\left(2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right)}\)
\(
=2.4 \times 10^{-5} \mathrm{~m}
\)

Hint

\(
\begin{aligned}
& Y=\frac{\text { stress }}{\text { strain }}=\frac{T / A}{l / L} \\
& \text { or, } \quad l=\frac{T L}{A Y} \\
& \text { or, elongation }=l=\frac{(100 \mathrm{~N}) \times(4 \cdot 5 \mathrm{~m})}{\left(\pi \times 9 \times 10^{-6} \mathrm{~m}^2\right) \times\left(4 \cdot 8 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right)} \\
& =3.32 \times 10^{-5} \mathrm{~m} \text {. } \\
& \text { Again, Poisson ratio }=\frac{\Delta d / d}{l / L}=\frac{(\Delta d) L}{l d} \\
& \text { or, } \quad 0.2=\frac{\Delta d \times 4.5 \mathrm{~m}}{\left(3.32 \times 10^{-5} \mathrm{~m}\right) \times\left(6 \times 10^{-3} \mathrm{~m}\right)} \\
& \text { or, } \quad \Delta d=\frac{0.2 \times 6 \times 3.32 \times 10^{-8} \mathrm{~m}}{4.5} \\
& =8.8 \times 10^{-9} \mathrm{~m} \text {. } \\
&
\end{aligned}
\)

Hint

The stress in the wire \(=\frac{\text { Tension }}{\text { Area of cross section }}\).
To avoid breaking, this stress should not exceed the breaking stress.
Let the tension in the wire be \(T\). The equations of motion of the two blocks are,
and
\(
T-10 \mathrm{~N}=(1 \mathrm{~kg}) a
\)
Eliminating \(a\) from these equations,
\(
\begin{aligned}
T & =(40 / 3) \mathrm{N} . \\
\text { The stress } & =\frac{(40 / 3) \mathrm{N}}{\pi r^2} .
\end{aligned}
\)
If the minimum radius needed to avoid breaking is \(r\),
\(
2 \times 10^9 \frac{\mathrm{N}}{\mathrm{m}^2}=\frac{(40 / 3) \mathrm{N}}{\pi r^2}
\)
Solving this,
\(
r=4.6 \times 10^{-5} \mathrm{~m}
\)

Hint

As the cross sections of the wires are equal and same tension exists in both, the stresses developed are equal. Let the original lengths of the steel wire and the copper wire be \(L_s\) and \(L_c\) respectively and the elongation in each wire be \(l\).
and
\(
\begin{aligned}
\frac{l}{L_s} & =\frac{\text { stress }}{2 \cdot 0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} \dots(i)\\
\frac{l}{L_c} & =\frac{\text { stress }}{1 \cdot 1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} \dots(ii)
\end{aligned}
\)
Dividing (ii) by (i),
\(
L_s / L_c=2 \cdot 0 / 1 \cdot 1=20: 11 .
\)

Hint

The situation is described in the figure below. As the \(1 \mathrm{~kg}\) mass is in equilibrium, the tension in the lower wire equals the weight of the load.

Thus \(\quad T_1=10 \mathrm{~N}\)
\(
\begin{aligned}
\text { Stress } & =10 \mathrm{~N} / 0.005 \mathrm{~cm}^2 \\
& =2 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2} .
\end{aligned}
\)
Longitudinal strain \(=\frac{\text { stress }}{Y}=\frac{2 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2}}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}=10^{-4}\).
Considering the equilibrium of the upper block, we can write,
\(
\begin{aligned}
T_2 & =20 \mathrm{~N}+T_1, \quad \text { or, } \quad T_2=30 \mathrm{~N} . \\
\text { Stress } & =30 \mathrm{~N} / 0.005 \mathrm{~cm}^2 \\
& =6 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2} .
\end{aligned}
\)
Longitudinal strain \(=\frac{6 \times 10^7 \mathrm{~N} \mathrm{~m}^{-2}}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}=3 \times 10^{-4}\).

Hint

The block \(R\) will descend vertically and the blocks \(P\) and \(Q\) will move on the frictionless horizontal table. Let the common magnitude of the acceleration be \(a\). Let the tensions in the wires \(A\) and \(B\) be \(T_A\) and \(T_B\) respectively.
Writing the equations of motion of the blocks \(P, Q\) and \(R\), we get,
\(
T_{\mathrm{A}}=(3 \mathrm{~kg}) a \dots(i)
\)
\(
T_B-T_A=(3 \mathrm{~kg}) a \dots(ii)
\)
and \((3 \mathrm{~kg}) g-T_B=(3 \mathrm{~kg}) a \dots(iii)\).
By (i) and (ii),
\(
T_B=2 T_A .
\)
By (i) and (iii),
\(
\begin{aligned}
& T_A+T_B=(3 \mathrm{~kg}) g=30 \mathrm{~N} \\
& \text { or, } \quad 3 T_A=30 \mathrm{~N} \\
& \text { or, } \quad T_A=10 \mathrm{~N} \text { and } T_B=20 \mathrm{~N} . \\
& \text { Longitudinal strain }=\frac{\text { Longitudinal stress }}{\text { Young modulus }} \\
& \text { Strain in wire } A=\frac{10 \mathrm{~N} / 0 \cdot 005 \mathrm{~cm}^2}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}=10^{-4} \\
& \text { and strain in wire } B=\frac{20 \mathrm{~N} / 0 \cdot 005 \mathrm{~cm}^2}{2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}=2 \times 10^{-4} .
\end{aligned}
\)

Hint

The volume of the wire is
\(
\begin{aligned}
V & =\left(3.0 \mathrm{~mm}^2\right)(50 \mathrm{~cm}) \\
& =\left(3.0 \times 10^{-6} \mathrm{~m}^2\right)(0.50 \mathrm{~m})=1.5 \times 10^{-6} \mathrm{~m}^3 .
\end{aligned}
\)
Tension (tensile Force) in the wire is
\(
\begin{aligned}
T & =m g \\
& =(2.1 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)=21 \mathrm{~N} .
\end{aligned}
\)
The stress \(=T / A = F / A\)
\(
=\frac{21 \mathrm{~N}}{3.0 \mathrm{~mm}^2}=7 \cdot 0 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2} \text {. }
\)
The strain \(=\) stress \(/ Y\)
\(
=\frac{7 \cdot 0 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}}{1 \cdot 9 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}}=3.7 \times 10^{-5} \text {. }
\)
The elastic potential energy of the wire is
\(
\begin{aligned}
U & =\frac{1}{2}(\text { stress) }(\text { strain) (volume) } \\
& =\frac{1}{2}\left(7 \cdot 0 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}\right)\left(3.7 \times 10^{-5}\right)\left(1.5 \times 10^{-6} \mathrm{~m}^3\right) \\
& =1 \cdot 9 \times 10^{-4} \mathrm{~J} .
\end{aligned}
\)

Hint

Forces acting on the block are (a) the tension \(T\) and (b) the weight \(W\). At the lowest point, the resultant force is \(T-W\) towards the centre. As the block is going in a circle, the net force towards the centre should be \(m v^2 / r\) with usual symbols. Thus,
or,
\(
\begin{aligned}
T-W & =m v^2 / r \\
T & =W+m v^2 / r \\
& =10 \mathrm{~N}+\frac{(1 \mathrm{~kg})\left(2 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{0 \cdot 2 \mathrm{~m}}=30 \mathrm{~N}
\end{aligned}
\)
We have
or,
\(
\begin{aligned}
Y & =\frac{T / A}{l / L} \\
l & =\frac{T L}{A Y} \\
& =\frac{30 \mathrm{~N} \times(20 \mathrm{~cm})}{\left(3 \times 10^{-6} \mathrm{~m}^2\right) \times\left(2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right)} \\
& =5 \times 10^{-5} \times 20 \mathrm{~cm}=10^{-3} \mathrm{~cm} .
\end{aligned}
\)

Hint

Consider a small length \(d x\) of the rod at a distance \(x\) from the fixed end. The part below this small element has length \(L-x\). The tension \(T\) of the rod at the element equals the weight of the rod below it.

\(
T=(L-x) \frac{W}{L}
\)
Elongation in the element is given by
\(
\begin{aligned}
\text { elongation } & =\text { original length } \times \text { stress } / Y \\
& =\frac{T d x}{A Y}=\frac{(L-x) W d x}{L A Y} .
\end{aligned}
\)
\(
\begin{aligned}
\text { The total elongation } & =\int_0^L \frac{(L-x) W d x}{L A Y} \\
& =\frac{W}{L A Y}\left(L x-\frac{x^2}{2}\right)_0^L=\frac{W L}{2 A Y} .
\end{aligned}
\)

Hint

Given:
Mass of the load \((\mathrm{m})=10 \mathrm{~kg}\)
Length of wire \((L)=3 \mathrm{~m}\)
Area of cross-section of the wire \((\mathrm{A})=4 \mathrm{~mm}^2=4.0 \times 10^{-6} \mathrm{~m}^2\)
Young’s modulus of the metal \(Y=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)

(a) Stress = F/A
\(
\begin{aligned}
& \begin{aligned}
\mathrm{F}= & \mathrm{mg} \\
& =10 \times 10=100 \mathrm{~N} \quad\left(\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\right) \\
\therefore \frac{F}{A} & =\frac{100}{4 \times 10^{-6}} \\
= & 2.5 \times 10^7 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\end{aligned}
\)

(b) Strain \(=\frac{\Delta L}{L}\)
Or,
Strain \(=\frac{\text { Stress }}{Y}\)
Strain \(=\frac{2.5 \times 10^7}{2 \times 10^{11}}\)
\(=1.25 \times 10^{-4}\)

(c) Let the elongation in the wire be \(\Delta L\).
\(
\begin{aligned}
& \text { Strain }=\frac{\Delta L}{L} \\
& \Rightarrow \Delta L=(\text { Strain }) \times L \\
& =1.25 \times 10^{-4} \times 3 \\
& =3.75 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

Given:
Radius of cylinder \((r)=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
Length of cylinder \((\mathrm{L})=2 \mathrm{~m}\)
Mass of the load \(=100 \mathrm{~kg}\)
Young’s modulus of the metal \(=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)

(a) Stress is given by : \(\frac{F}{A}\)
Here, \(\mathrm{F}\) is the force given by \(\mathrm{mg}=100 \times 10=1000 \mathrm{~N}\) ( Taking \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )
\(A\) is the area of cross-section \(=\pi r^2=4 \pi \times 10^{-4} \mathrm{~m}^2\)
\(
\begin{aligned}
& \Rightarrow \text { Stress } =\frac{\mathrm{mg}}{A} \\
& =\frac{(100 \times 10)}{\left(4 \pi \times 10^{-4}\right)} \\
& =7.96 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

(b) Strain is given by:
\(
\begin{aligned}
& \text { Strain }=\frac{\text { Stress }}{Y}=\frac{\left(7.96 \times 10^5\right)}{\left(2 \times 10^{11}\right)} \\
& =4 \times 10^{-6}
\end{aligned}
\)

(c) Compression of the cylinder:
\(
\begin{aligned}
\Delta L & =\text { strain } \times L \\
& =4 \times 10^{-6} \times 2=8 \times 10^{-6} \mathrm{~m}
\end{aligned}
\)

Hint

Given:
Elastic limit of steel \(\frac{F}{A}=8 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)
Young’s modulus of steel \(Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Length of steel wire \(L=\frac{1}{2} \mathrm{~m}=0.5 \mathrm{~m}\)
The elastic limit of steel indicates the maximum pressure that steel can bear.
Let the maximum elongation of steel wire be \(\Delta L\).
\(
\begin{aligned}
& Y=\frac{F}{A} \frac{L}{\Delta L} \\
& \Rightarrow \Delta L=\frac{F L}{A Y} \\
& \Rightarrow \Delta L=\frac{8 \times 10^5 \times(0.5)}{2 \times 10^{11}} \\
& =2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~mm}
\end{aligned}
\)
Hence, the required elongation of steel wire is \(2 \mathrm{~mm}\).

Hint

Given:
Young’s modulus of steel \(=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)
Young’s modulus of copper \(=1.3 \times 1011 \mathrm{~N} \mathrm{~m}^{-2}\)
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
As per the question:
\(
\begin{aligned}
& L_{\text {steel }}=L_{\text {Cu }} \\
& A_{\text {steel }}=A_{\mathrm{Cu}} \\
& F_{\text {Cu }}=F_{\text {Steel }}
\end{aligned}
\)
Here: \(\mathrm{L}_{\text {steel }}\) and \(\mathrm{L}_{\text {Cu }}\) denote the lengths of steel and copper wires, respectively.
\(\mathrm{A}_{\text {steel }}\) and \(\mathrm{A}_{\mathrm{Cu}}\) denote the cross-sectional areas of steel and copper wires, respectively.
\(\mathrm{F}_{\text {steel }}\) and \(\mathrm{F}_{\mathrm{Cu}}\) denote the tension of steel and cooper wires, respectively.

(a)
\(
\frac{\text { Stress of } \mathrm{Cu}}{\text { Stress of Steel }}=\frac{F_{\mathrm{Cu}}}{A_{\mathrm{Cu}}} \frac{A_{\text {Steel }}}{F_{\text {Steel }}}=1
\)

(b)
\(
\begin{aligned}
& \frac{\text { Strain of } \mathrm{Cu}}{\text { Strain of steel }}=\frac{\frac{\Delta L_{\text {Steel }}}{L_{\text {Steel }}}}{\frac{\Delta L_{\mathrm{cu}}}{L_{\mathrm{cu}}}}=\frac{F_{\text {Steel }} L_{\text {Steel }} A_{\mathrm{cu}} Y_{\mathrm{cu}}}{A_{\text {Steel }} Y_{\text {Steel }} F_{\mathrm{cu}} L_{\mathrm{cu}}} \\
& \left(\text { Using } \frac{\Delta L}{L}=\frac{F}{A Y}\right) \\
& \Rightarrow \frac{\text { Strain of } \mathrm{Cu}}{\text { Strain of steel }}=\frac{Y_{\mathrm{cu}}}{Y_{\text {Steel }}}=\frac{1.3 \times 10^{11}}{2 \times 10^{11}} \\
& \Rightarrow \frac{\text { Strain of Cu }}{\text { Strain of steel }}=\frac{13}{20} \\
& \Rightarrow \frac{\text { Strain of steel }}{\text { Strain of Cu }}=\frac{20}{13}
\end{aligned}
\)
Hence, the required ratio is \(20: 13\).

Hint

Given that both wires are of equal length and equal cross-sectional area, the block applies equal tension on both of them.
\(
\begin{aligned}
& \therefore L_{\text {steel }}=L_{\mathrm{Cu}} \\
& A_{\text {steel }}=A_{\mathrm{Cu}} \\
& F_{\text {Cu }}=F_{\text {Steel }} \\
& \frac{\text { Strain of } \mathrm{Cu}}{\text { Strain of steel }}=\frac{\frac{\Delta L_{\text {Steel }}}{L_{\text {Steel }}}}{\frac{\Delta L_{\mathrm{cu}}}{L_{\mathrm{cu}}}}=\frac{F_{\text {Steel }} L_{\text {Steel }} A_{\mathrm{cu}} Y_{\mathrm{cu}}}{A_{\text {Steel }} Y_{\text {Steel }} F_{\mathrm{cu}} L_{\mathrm{cu}}} \\
& \left(\text { Using } \frac{\Delta L}{L}=\frac{F}{A Y}\right) \\
& \Rightarrow \frac{\text { Strain of } \mathrm{Cu}}{\text { Strain of steel }}=\frac{Y_{\text {cu }}}{Y_{\text {Steel }}}=\frac{1.3 \times 10^{11}}{2 \times 10^{11}} \\
& \Rightarrow \frac{\text { Strain of steel }}{\text { Strain of Cu }}=\frac{20}{13}=1.54
\end{aligned}
\)

Hint

Given: Breaking stress of wire \(=8 \times 10^8 \mathrm{~N} / \mathrm{m}^2\)
Area of cross-section of upper wire \(\left(A_{\mathrm{u}}\right)=0.006 \mathrm{~cm}^2=6 \times 10^{-7} \mathrm{~m}\)
Area of cross-section of lower wire \(\left(A_l\right)=0.003 \mathrm{~cm}^2=3 \times 10^{-7} \mathrm{~m}\) \(m_1=10 \mathrm{~kg}, m_2=20 \mathrm{~kg}\)
(a)
Tension in lower wire
\(
T_1=\mathrm{m}_1 \mathrm{~g}+\mathrm{w}
\)
Here : \(g\) is the acceleration due to gravity, \(w\) is the load
\(\therefore\) Stress in lower wire
\(
\begin{aligned}
& =\frac{\mathrm{T}_1}{\mathrm{~A}_1}=\frac{\mathrm{m}_1 \mathrm{~g}+\mathrm{w}}{\mathrm{A}_1} \\
& \Rightarrow \frac{\mathrm{m}_1 \mathrm{~g}+\mathrm{w}}{\mathrm{A}_{\mathrm{l}}}=8 \times 10^8 \\
& \Rightarrow \mathrm{w}=\left[\left(8 \times 10^8\right) \times\left(3 \times 10^{-7}\right)\right]-100 \\
& \Rightarrow \mathrm{w}=140 \mathrm{~N} \text { or } 14 \mathrm{~kg}
\end{aligned}
\)
Now, tension in upper wire
\(T_2=m_1 g+m_2 g+w\)
\(\therefore\) Stress in upper wire
\(
\begin{aligned}
& =\frac{T_u}{A_u}=\frac{m_2 g+m_1 g+w}{A_u} \\
& \Rightarrow \frac{m_2 g+m_1 g+w}{A_u}=8 \times 10^8 \\
& \Rightarrow w=180 \mathrm{~N} \text { or } 18 \mathrm{~kg}
\end{aligned}
\)
For the same breaking stress, the maximum load that can be put is \(140 \mathrm{~N}\) or \(14 \mathrm{~kg}\). The lower wire will break first if the load is increased.
(b)
If \(\mathrm{m}_1=10 \mathrm{~kg}\) and \(\mathrm{m}_2=36 \mathrm{~kg}\)
Tension in lower wire \(T_1=m_1 g+w\)
\(\therefore\) Stress in lower wire:
\(
\begin{aligned}
& \Rightarrow \frac{T_1}{A_l}=\frac{m_1 g+w}{A_l}=8 \times 10^5 \\
& \Rightarrow \mathrm{w}=140 \mathrm{~N}
\end{aligned}
\)
Now, tension in upper wire
\(
\begin{aligned}
& \mathrm{T}_2=\mathrm{m}_1 \mathrm{~g}+\mathrm{m}_2 \mathrm{~g}+\mathrm{w} \\
& \therefore \text { Stress in upper wire: } \\
& \Rightarrow \frac{\mathrm{T}_{\mathrm{u}}}{\mathrm{A}_{\mathrm{u}}}=\frac{\mathrm{m}_2 \mathrm{~g}+\mathrm{m}_1 \mathrm{~g}+\mathrm{w}}{\mathrm{A}_{\mathrm{u}}}=8 \times 10^5 \\
& \Rightarrow \mathrm{w}=20 \mathrm{~N} = 2 \mathrm{~kg}
\end{aligned}
\)
For the same breaking stress, the maximum load that can be put is \(20 \mathrm{~N}\) or \(2 \mathrm{~kg}\). The upper wire will break first if the load is increased.

Hint

Given:
Force \((\mathrm{F})\) applied by two persons on the rope \(=100 \mathrm{~N}\)
Original length of rope \(\mathrm{L}=2 \mathrm{~m}\)
Extension in the rope \(\Delta L=0.01 \mathrm{~m}\)
Area of cross – section of the rope \(A=2 \times 10^{-4}\)
We know that :
Young’s modulus \(Y=\frac{F}{A} \times \frac{L}{\Delta L}\)
\(
\begin{aligned}
& =\frac{100}{2 \times 10^{-4}} \times \frac{2}{0.01} \\
& \Rightarrow Y=1 \times 10^8 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Hence, the required Young’s modulus for the rope is \(1 \times 10^8 \mathrm{~N} / \mathrm{m}^2\).

Hint

Given:
Cross-sectional area of steel \(\operatorname{rod} \mathrm{A}=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2\)
Length of steel \(\operatorname{rod} \mathrm{L}=2 \mathrm{~m}\)
Compression during night hours \(\Delta \mathrm{L}=0.1 \mathrm{~cm}=10^{-3} \mathrm{~m}\)
Young modulus of steel \(Y=1.9 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)
Let the tension developed at night be \(\mathrm{F}\).
\(
\begin{aligned}
& Y=\frac{F}{A} \times \frac{L}{\Delta L} \\
& \Rightarrow F=\frac{Y A \Delta L}{L} \\
& =\frac{1.9 \times 10^{11} \times 4 \times 10^{-4} \times 10^{-3}}{2} \\
& =3.8 \times 10^4 N
\end{aligned}
\)
\(\therefore\) Required tension developed in steel rod during night hours \(=3.8 \times 10^4 \mathrm{~N}\).

Hint

Given:
Force \((F)=m_2 g / 2\)
Area of cross-section of the string \(=\mathrm{A}\)
Young’s modulus \(=Y\)
Let \(a\) be the acceleration produced in block \(\mathrm{m}_2\) in the downward direction and \(\mathrm{T}\) be the tension in the string.

From the free body diagram:
\(
\begin{aligned}
& m_2 \mathrm{~g}-\mathrm{T}=\mathrm{m}_2 \mathrm{a} \ldots \text { (i) } \\
& \mathrm{T}-\mathrm{F}=\mathrm{m}_1 \mathrm{a} \ldots \text { (ii) }
\end{aligned}
\)
From equations (i) and (ii), we get:
\(
\begin{aligned}
& a=\frac{m_2 g-F}{m_1+m_2} \\
& \text { Applying } F=\frac{m_2 g}{2} \\
& \Rightarrow a=\frac{m_2 g}{2\left(m_1+m_2\right)}
\end{aligned}
\)
Again, \(\mathrm{T}=\mathrm{F}+\mathrm{m}_1 \mathrm{a}\)
On applying the values of \(F\) and \(a\), we get:
\(
\Rightarrow T=\frac{m_2 g}{2}+m_1 \frac{m_2 g}{2\left(m_1+m_2\right)}
\)
We know that:
\(
\begin{aligned}
& \mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}} \\
& \Rightarrow \text { Strain }=\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}} \\
& \Rightarrow \text { Strain }=\frac{\left(\mathrm{m}_2^2+2 \mathrm{~m}_1 \mathrm{~m}_2\right) \mathrm{g}}{2\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{AY}} \\
& =\frac{\mathrm{m}_2 g\left(2 \mathrm{~m}_1+\mathrm{m}_2\right)}{2 \mathrm{AY}\left(\mathrm{m}_1+\mathrm{m}_2\right)}
\end{aligned}
\)

Hint

Given:
Mass of sphere \((m)=20 \mathrm{~kg}\)
Length of metal wire \((\mathrm{L})=4 \mathrm{~m}\)
Diameter of wire \((\mathrm{d}=2 \mathrm{r})=1 \mathrm{~mm}\)
\(
\Rightarrow \mathrm{r}=5 \times 10^{-4} \mathrm{~m}
\)
Young’s modulus of the metal wire \(=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)
Tension in the wire in equilibrium \(=T\)
\(
\mathrm{T}=\mathrm{mg}
\)
When it is moved at an angle \(\theta\) and released, let the tension at the lowest point be \(\mathrm{T}^{\prime}\).
\(
\Rightarrow T^{\prime}=m g+\frac{m v^2}{r}
\)
The change in tension is due to the centrifugal force.
\(
\begin{aligned}
& \therefore \Delta \mathrm{T}=\mathrm{T}^{\prime}-\mathrm{T} \\
& \Delta \mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}} \ldots \text { (1) }
\end{aligned}
\)
Now, using the work energy principle:
\(
\begin{aligned}
& \frac{1}{2} \mathrm{mv}^2-0=\operatorname{mgr}(1-\cos \theta) \\
& \Rightarrow \mathrm{v}^2=2 \mathrm{gr}(1-\cos \theta) \ldots(2)
\end{aligned}
\)
Applying the value of \(v^2\) in (1):
\(
\begin{aligned}
& \Delta T=\frac{\mathrm{m}[2 \mathrm{gr}(1-\cos \theta)]}{r} \\
& =2 \mathrm{mg}(1-\cos \theta)
\end{aligned}
\)
Now, \(\mathrm{F}=\Delta \mathrm{T}\)
\(
\begin{aligned}
& \text { Also, } F=\frac{Y A \Delta L}{L} \\
& \Rightarrow \frac{Y A \Delta L}{L}=2 \mathrm{mg}(1-\cos \theta) \\
& \Rightarrow \cos \theta=1-\frac{Y A \Delta L}{L(2 \mathrm{mg})} \\
& \Rightarrow \cos \theta=1-\left[\frac{2 \times 10^{11} \times 4 \times 3.14 \times(5)^2 \times 10^{-8} \times 2 \times 10^{-3}}{4 \times 2 \times 20 \times 10}\right] \\
& \Rightarrow \cos \theta=0.80 \\
& \text { Or, } \theta=36.4^{\circ}
\end{aligned}
\)

Hint

Given: Original length of steel wire \((\mathrm{L})=1 \mathrm{~m}\)
Area of cross-section \((\mathrm{A})=4.00 \mathrm{~mm}^2=4 \times 10^{-2} \mathrm{~cm}^2\)
Load \(=2.16 \mathrm{~kg}\)
Young’s modulus of steel \((Y)=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} \mathrm{~s}^{-2}\)
Let \(\mathrm{T}\) be the tension in the string after the load is suspended and \(\theta\) be the angle made by the string with the vertical, as shown in the figure:
\(
\cos \theta=\frac{x}{\sqrt{x^2+l^2}}=\frac{x}{l}\left\{1+\frac{x^2}{l^2}\right\}^{-1 / 2}
\)
Expanding the above equation using the binomial theorem:
\(\cos \theta=\frac{x}{l}\left\{1-\frac{1}{2} \frac{x^2}{l^2}\right\}\) (neglecting the higher order terms )
Since \(\mathrm{x}<<l, \frac{x^2}{{l}^2}\) can be neglected.
\(
\Rightarrow \cos \theta=\frac{x}{{l}}
\)
Increase in length:
\(
\begin{aligned}
& \Delta \mathrm{L}=(\mathrm{AC}+\mathrm{CB})-\mathrm{AB} \\
& \mathrm{AC}=\left({l}^2+\mathrm{x}^2\right)^{1 / 2} \\
& \Delta \mathrm{L}=2\left({l}^2+\mathrm{x}^2\right)^{1 / 2}-2 {l}
\end{aligned}
\)
We know that : \(Y=\frac{F}{A} \frac{L}{\Delta L}\)
\(
\Rightarrow 2 \times 10^{12}=\frac{T \times 100}{\left(4 \times 10^{-2}\right) \times\left[2\left(50^2+x^2\right)^{1 / 2}-100\right]}
\)
From the free body diagram:
\(
\begin{aligned}
& 2 T \cos \theta=\mathrm{mg} \\
& 2 T\left(\frac{x}{50}\right)=2.16 \times 10^3 \times 980 \\
& \Rightarrow \frac{2 \times\left(2 \times 10^{12}\right) \times\left(4 \times 10^{-2}\right) \times\left[2\left(50^2+x^2 \frac{1}{2}\right)-100\right] x}{100 \times 50}=(2.16) \times 10^3 \times 980
\end{aligned}
\)
On solving the above equation, we get \(x=1.5 \mathrm{~cm}\).
Hence, the required vertical depression is \(1.5 \mathrm{~cm}\).

Hint

Given:
Cross-sectional area of copper wire \(A=0.01 \mathrm{~cm}^2=10^{-6} \mathrm{~m}^2\)
Applied tension \(T=20 \mathrm{~N}\)
Young modulus of copper \(Y=1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Poisson ratio \(\sigma=0.32\)
We know that: \(Y=\frac{F L}{A \Delta L}\)
\(
\begin{aligned}
& \Rightarrow \frac{\Delta L}{L}=\frac{F}{A Y} \\
& =\frac{20}{10^{-6} \times 1.1 \times 10^{11}}=18.18 \times 10^{-5}
\end{aligned}
\)
Poisson’s ratio, \(\sigma=\frac{\frac{\Delta d}{d}}{\frac{\Delta L}{L}}=0.32\)
Where \(d\) is the transverse length
So, \(\frac{\Delta d}{d}=(0.32) \times \frac{\Delta L}{L}\)
\(=0.32 \times(18.18) \times 10^{-5}=5.81 \times 10^{-5}\)
Again , \(\frac{\Delta A}{A}=\frac{2 \Delta r}{r}=\frac{2 \Delta d}{d}\)
\(
\begin{aligned}
& \Rightarrow \Delta A=\frac{2 \Delta d}{d} A \\
& \Rightarrow \Delta A=2 \times\left(5.8 \times 10^{-5}\right) \times(0.01) \\
& =1.164 \times 10^{-6} \mathrm{~cm}^2
\end{aligned}
\)
Hence, the required decrease in the cross-sectional area is \(1.164 \times 10^{-6} \mathrm{~cm}^2\)

Hint

Bulk modulus of water \((B)=2.1 \times 10^9 \mathrm{Nm}^{-2}\)
In order to decrease the volume \((\mathrm{V})\) of a water sample by \(0.01 \%\), let the increase in pressure be \(\mathrm{P}\).
\(
\begin{aligned}
& \frac{V \times 0.01}{100}=\Delta V \\
& \Rightarrow \frac{\Delta V}{V}=10^{-4}
\end{aligned}
\)
From \(B=\frac{P V}{\Delta V}\), we have :
\(
\begin{aligned}
& \Rightarrow P=B\left(\frac{\Delta V}{V}\right) \\
& =2.1 \times 10^9 \times 10^{-4} \\
& =2.1 \times 10^5 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Hence, the required increase in pressure is \(2.1 \times 10^5 \mathrm{Nm}^{-2}\).

Hint

Given :
Bulk modulus of water \(B=2 \times 10^9 \mathrm{~N} / \mathrm{m}^2\)
Depth \((d)=400 \mathrm{~m}\)
Density of water at the surface \(\left(\rho_0\right)=1030 \mathrm{~kg} / \mathrm{m}^3\)
We know that:
Density at surface \(\rho_0=\frac{\mathrm{m}}{V_0}\)
Density at depth \(\rho_d=\frac{\mathrm{m}}{V_d}\)
\(
\Rightarrow \frac{\rho_d}{\rho_0}=\frac{V_0}{V_d} \ldots \text { (i) }
\)
Here: \(\rho_d=\) density of water at a depth
\(m=\) mass
\(V_0=\) volume at the surface
\(V_{d}=\) volume at a depth
Pressure at a depth \({d}=\rho_0 \mathrm{gd}\)
Acceleration due to gravity \(\mathrm{g}=10 \mathrm{~ms}^2\)
Volume strain \(=\frac{V_0-V_d}{V_0}\)
\(
\begin{aligned}
& B=\frac{\text { Pressure }}{\text { Volume strain }} \\
& \Rightarrow B=\frac{\rho_0 \mathrm{gd}}{\left(\frac{V_0-V_d}{V_0}\right)} \\
& \Rightarrow 1-\frac{V_d}{V_0}=\frac{\rho_0 \mathrm{gd}}{B} \\
& \Rightarrow \frac{V_d}{V_0}=\left(1-\frac{p_0 \mathrm{gd}}{B}\right) \dots(ii)
\end{aligned}
\)
Using equations (i) and (ii), we get:
\(
\begin{aligned}
& \frac{\rho_d}{\rho_0}=\frac{1}{\left(1-\frac{\rho_0 g \mathrm{~d}}{B}\right)} \\
& \Rightarrow \rho_d=\frac{1}{\left(1-\frac{\rho_0 \mathrm{gd}}{B}\right)} \rho_0 \\
& \Rightarrow \rho_d=\frac{1030}{\left(1-\frac{1030 \times 10 \times 400}{2 \times 10^9}\right)} \approx 1032 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
\)
Change in density \(=\rho_d-\rho_0\)
\(
=1032-1030=2 \mathrm{~kg} / \mathrm{m}^3
\)
Hence, the required density at a depth of \(400 \mathrm{~m}\) below the surface is \(2 \mathrm{~kg} / \mathrm{m}^3\).

Hint

Given:
Face area of steel plate \(A=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2\)
Thickness of steel plate \(d=0.5 \mathrm{~cm}=0.5 \times 10^{-2} \mathrm{~m}\)
Applied force on the upper surface \(F=10 \mathrm{~N}\)
Rigidity modulus of steel \(=8.4 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}\)
Let \(\theta\) be the angular displacement.
Rigidity modulus \(\mathrm{m}=\frac{F}{A \theta}\)
\(
\begin{aligned}
& \Rightarrow m=\left(\frac{10}{4 \times 10^{-4} \theta}\right) \\
& \Rightarrow \theta=\frac{10}{4 \times 10^{-4} \times 8.4 \times 10^{10}} \\
& =0.297 \times 10^{-6}
\end{aligned}
\)
\(\therefore\) Lateral displacement of the upper surface with respect to the lower surface \(=\theta \times d\)
\(
\Rightarrow(0.297) \times 10^{-6} \times(0.5) \times 10^{-2}
\)
\(
\Rightarrow 1.5 \times 10^{-9} \mathrm{~m}
\)
Hence, the required lateral displacement of the steel plate is \(1.5 \times 10^{-9} \mathrm{~m}\).

Hint

\(
F_1=500 \mathrm{~N}
\)
Let the required breaking force on the \(2 \mathrm{~cm}\) wire be \(F\).
Breaking stress in \(1 \mathrm{~cm}\) wire \(=\frac{F_1}{A_1}=\frac{500}{\pi\left(\frac{0.01}{2}\right)^2}\)
Breaking stress in \(2 \mathrm{~cm}\) wire \(=\frac{F_2}{A_2}=\frac{F_2}{\pi\left(\frac{0.02}{2}\right)^2}\)
The breaking stress is the same for a material.
\(
\begin{aligned}
& \Rightarrow \frac{500}{\pi\left(\frac{0.01}{2}\right)^2}=\frac{F_2}{\pi\left(\frac{0.02}{2}\right)^2} \\
& =>F_2=2000 \mathrm{~N}
\end{aligned}
\)

Hint

Breaking stress depends upon the intermolecular/inter-atomic forces of attraction within materials. In other words, it depends upon the material of the wire.

Hint

The stress at which the wire can break or rupture is called breaking stress. This is an intrinsic property of the material and depends only on the nature of the material.
Here, the breaking stress of the wire corresponds to the weight of $20\text{kg}$ mass. As the wire is cut into two equal parts, the nature of the material does not change. Hence, its breaking stress will not change. i.e each part of the wire can sustain a weight of $20\text{kg}$ mass.

Hint

Let the Young’s modulus of the wire’s material be \(Y\).
Here:
\(
\begin{aligned}
& \text { Force }=\mathrm{F} \\
& A_1=\pi r^2 \\
& L_1=l \\
& A_2=\pi\left(\frac{r}{2}\right)^2=\frac{\pi r^2}{4} \\
& \mathrm{~L}_2=2l
\end{aligned}
\)
Let the elongation in \(A\) be \(x\) and that in \(B\) be \(y\).
Since the Young’s modulus for both the wires is the same:
\(
\begin{aligned}
& Y=\frac{\frac{F}{A_1}}{\frac{x}{l}}=\frac{\frac{F}{A_2}}{\frac{y}{2 l}} \\
& \Rightarrow \frac{x}{y}=\frac{A_2}{2 A_1} \\
& \Rightarrow \frac{x}{y}=\frac{1}{8}
\end{aligned}
\)