Elasticity

$Y=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$

$ \Rightarrow \quad \Delta L=\frac{F L}{A Y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, $m=\rho V=\rho A L$

$ \therefore \quad L=\frac{m}{\rho A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

∴ From Eqs. (i) and (ii), we get

$ \begin{aligned} \Delta L & =\frac{F \cdot \frac{m}{\rho A}}{A Y}=\frac{F m}{\rho A^2 Y} \\ \Rightarrow \quad \Delta L & \propto \frac{1}{A^2} \\ \Rightarrow \frac{(\Delta L)_1}{(\Delta L)_2} & =\frac{A_2^2}{A_1^2}=\frac{\left(l^2\right)^2}{\left(\pi r^2\right)^2} \\ & =\frac{\left(1 \times 10^{-2}\right)^4}{\pi^2 \times\left(0.5 \times 10^{-2}\right)^4}=\frac{16}{\pi^2} \end{aligned} $

$ \begin{aligned} & \text {Work }=\text { Elastic potential energy } \\ & =\frac{1}{2} \times Y \times(\text { strain })^2 \times \text { volume } \\ & \Rightarrow W=\frac{1}{2} \times 2 \times 10^{11} \times\left(10^{-3}\right)^2 \times \frac{2.96}{7.4 \times 10^3} \\ & \Rightarrow W=40 \mathrm{~J} \text { or } 0.04 \mathrm{~kJ} \end{aligned} $

$ \begin{aligned} & \text { } A=\pi r^2 \\ & \therefore \frac{d A}{d r}=2 \pi r \Rightarrow d A=2 \pi r d r \\ & \Delta A=2 \pi r \Delta r \end{aligned} $

Thus, $\frac{\Delta A}{A}=\frac{2 \Delta r}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

∴ Poisson's ratio,

$ \begin{aligned} \sigma & =-\frac{\text { Lateral strain }}{\text { Longitudinal strain }} \\ & =\frac{-\Delta r / r}{\Delta l / l} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{\Delta r}{r}=-\sigma \frac{\Delta l}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \therefore & Y=\frac{\text { stress }}{\text { strain }}=\frac{F / A}{\Delta l / l} \\ \Rightarrow & \frac{\Delta l}{l}=\frac{F}{A Y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

∴ From Eq. (ii) and (iii), we have,

$ \frac{\Delta r}{r}=\frac{-\sigma F}{A Y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Now, from Eq. (i) and (ivq), we get

$ \begin{aligned} & \frac{\Delta A}{A}=2\left(-\frac{\sigma F}{A Y}\right) \Rightarrow F=\frac{-Y \Delta A}{2 \sigma} \\ \therefore & |F|=\frac{Y \Delta A}{2 \sigma} \end{aligned} $

The energy stored per unit volume in a stretched rod

$ \begin{aligned} u & =\frac{1}{2} \times \text { stress × strain } \\ & =\frac{1}{2} \times \text { stress } \times \frac{\text { stress }}{Y} \\ & =\frac{1}{2} \times \frac{(\text { stress })^2}{Y}=\frac{1}{2} \times \frac{\left(\frac{F}{A}\right)^2}{Y} \\ & =\frac{1}{2} \times \frac{\left(\frac{9 \times 10^4}{3 \times 10^{-4}}\right)^2}{2 \times 10^{11}}=\frac{1}{2} \times \frac{9 \times 10^{16}}{2 \times 10^{11}} \\ & =2.25 \times 10^5 \mathrm{Jm}^{-3} \end{aligned} $

The mass of a wire is $m=\rho A L$

Since, the material is the same, the density $\rho$ is constant for both wires.

$ \begin{aligned} \frac{m_A}{m_B} & =\frac{\rho A_A L_A}{\rho A_B L_B} \Rightarrow \frac{m_A}{m_B}=\frac{A_A L_A}{A_B L_B} \\ \frac{2}{3} & =\frac{1}{2} \times \frac{L_A}{L_B} \\ \frac{L_A}{L_B} & =\frac{2}{3} \times 2=\frac{4}{3} \end{aligned} $

The ratio of elongations is

$ \frac{\Delta L_A}{\Delta L_B}=\frac{\frac{F L_A}{A_A Y}}{\frac{F L_B}{A_B Y}}=\frac{L_A}{A_A} \times \frac{A_B}{L_B} $

Substitute $\frac{L_A}{L_B}=\frac{4}{3}$ and $\frac{A_B}{A_A}=\frac{2}{1}$

$ \frac{\Delta L_A}{\Delta L_B}=\frac{4}{3} \times \frac{2}{1}=\frac{8}{3} $

The ratio of the elongations of the wires $A$ and $B$ is $\frac{8}{3}$.

$ \begin{aligned} B & =\frac{F}{A} \times \frac{V}{\Delta V} \\ B & =\frac{P V}{\Delta V} \end{aligned} $

$ \begin{aligned}Here,\,\, & V=\frac{4}{3} \pi R^3 \\ & \Delta V=4 \pi R^2 \Delta R \end{aligned} $

$ \begin{aligned} \therefore \quad B & =P \times \frac{4}{3} \pi R^3 \times \frac{1}{4 \pi R^2 \Delta R} \\ B & =\frac{P R}{3 \Delta R} \\ \Delta R & =\frac{P R}{3 B}=\frac{10^7 \times 36 \times 10^{-2}}{3 \times 60 \times 10^9} \\ \Delta R & =2 \times 10^{-5} \mathrm{~m} \\ & =2 \times 10^{-3} \mathrm{~cm} \end{aligned} $

Gravitational force acting on the pendulum,

$ F_{g}=m g $

where, $ m= $ mass of bob

Centripetal force acting on the pendulum,

$ F_{c}=\frac{m v^{2}}{R} $

where, $ v= $ speed of bus and $ R= $ radius of circular path

The net force experienced by the wire is the resultant of the gravitational force and the centripetal force. It can be determined by vector addition.

$ \begin{aligned} F_{\mathrm{net}} & =\sqrt{F_{g}^{2}+F_{c}^{2}} \\\\ & =\sqrt{(m g)^{2}+\left(\frac{m v^{2}}{R}\right)^{2}}=m \sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}} \\\\ & =\frac{m}{R} \sqrt{g^{2} R^{2}+v^{4}} \end{aligned} $

The elongation $ (\Delta L) $ in the wire,

$ \Delta L=\frac{F \cdot L}{A \cdot Y} $

where, $ F= $ net force, $ L= $ original length of wire, $ A= $ area of cross-section of the wire and $ Y= $ Young's modulus.

After putting values,

$ \Delta L=\frac{m \cdot L}{R A Y} \sqrt{g^{2} R^{2}+v^{4}} $

Given the following parameters:

Cross-sectional area: $ A = 10^{-6} \text{ m}^2 $

Elongation: $ \frac{\Delta l}{l} = 0.1\% = \frac{0.1}{100} $

Tension: $ T = 1000 \text{ N} $

We need to find the Young's modulus $ Y $ of the material of the wire. The formula for Young's modulus is:

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

Where stress is defined as:

$ \text{Stress} = \frac{\text{Tension}}{\text{Area}} $

And strain is the elongation:

$ \text{Strain} = \frac{\Delta l}{l} $

Substituting the given values into the formulas:

$ \text{Stress} = \frac{1000}{10^{-6}} $

$ \text{Strain} = \frac{0.1}{100} $

Now, calculating Young's modulus:

$ Y = \frac{1000}{10^{-6}} \times \frac{100}{0.1} $

Simplifying:

$ Y = 10^{12} \text{ N/m}^2 $

Therefore, the Young's modulus of the material of the wire is $ 10^{12} \text{ N/m}^2 $.

Mass of block, $ m=2 \mathrm{~kg} $

Length of metal wire, $ L=2 \mathrm{~m} $ At top (highest) point, tension $ T=0 $ So, the velocity required (critical) at bottom point to complete the revolution

$ v=\sqrt{5 g L} $

Tension at lowest point (where elongation is maximum)

$ \begin{aligned} T & =m g+\frac{m v^{2}}{L} \\\\ & =m g+\frac{m(5 g) L}{L} \\\\ T & =6 m g=6 \times 2 \times 10 \\\\ \text { Tension } & =120 \mathrm{~N} \\\\ \text { Strain } & =\frac{\Delta L}{L}=\frac{2 \times 10^{-3}}{2}=10^{-3} \\\\ \text { Stress } & =\frac{T}{A}=\frac{120}{1 \times 10^{-6}} \\\\ & =1.2 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2} \end{aligned} $

$ \begin{aligned} \text { Young's modulus } & =\frac{\text { Stress }}{\text { Strain }}=\frac{1.2 \times 10^{8}}{10^{-3}} \\\\ & =1.2 \times 10^{11} \mathrm{Nm}^{-2} \end{aligned} $

Let original length of the spring be $l \mathrm{~m}$ and Young's modulus be

$ Y=\frac{F \cdot l}{A \cdot \Delta l} $

Now, when $F=6 \mathrm{~N}, \Delta l=(P-l)$, then

$ Y=\frac{6 l}{A(P-I)} \quad \ldots \text { (i) } $

and when, $F=10 \mathrm{~N}$ and $\Delta l=(Q-l)$, then

$ Y=\frac{8 l}{A(Q-l)} \quad \ldots \text { (ii) } $

From Eqs. (i) and (ii), we get

$ \begin{array}{rlrl} & \frac{6 l}{A(P-l)} & =\frac{8 l}{A(Q-l)} \\\\ \Rightarrow & 6(Q-l) & =8(P-l) \\\\ & 6 Q-6 l & =8 P-8 l \\\\ \Rightarrow & -6 l+8 l & =8 P-6 Q \\\\ \Rightarrow & 2 l & =8 P-6 Q \\\\ \Rightarrow & l & =4 P-3 Q \end{array} $

Using Hooke's law,

$ F=k x $

The spring constant ( $k$ ) for material is given by

$ k=\frac{Y A}{L} $

Let denote the extension as $x_s$ and $x_C$

$ \begin{aligned} x_S+x_C & =1.05 \mathrm{~mm} \\\\ & =1.05 \times 10^{-3} \mathrm{~m} \end{aligned} $

Forces in both wires,

$ F=k_S x_S=k_C x_C $

Here, $\frac{Y_s A}{L_S} x_s=\frac{Y_c A}{L_C} x_C$

Let $x_s$ to $x_c$,

$ \frac{x_s}{x_c}=\frac{Y_c L_s}{Y_s L_c} $

$ \frac{x_S}{x_C}=\frac{1.1 \times 10^{11} \times 3}{2 \times 10^{11} \times 2.2}=\frac{3}{4} $

$ \begin{aligned} & \text { Thus, } \\\\ & \begin{aligned} \frac{3}{4} x_C+x_C & =1.05 \times 10^{-3} \\\\ \frac{7}{4} x_C & =1.05 \times 10^{-3} \\\\ x_C & =0.6 \times 10^{-3} \mathrm{~m} \end{aligned} \end{aligned} $

For $x_S$,

$ \begin{aligned} & x_s=\frac{3}{4} \times 0.6 \times 10^{-3} \\\\ & x_s=0.45 \times 10^{-3} \mathrm{~m} \end{aligned} $

Therefore, using the spring constant for steel,

$ \begin{aligned} & F=\frac{Y_S A}{L_S} x_S \\\\ & F=\frac{2 \times 10^{11} \times 6 \times 0.45}{3 \times 10^9} \\\\ & F=\frac{2 \times 6 \times 0.45}{3} \times 10^2 \\\\ & F=180 \mathrm{~N} \end{aligned} $

Therefore, the load applied is 180 N .

Given,

Ratio of area of cross-section of three wire $\left(A_1: A_2: A_3\right)=1: 2: 3$

Ratio of Young's modulus

$ \left(Y_1: Y_2: Y_3\right)=3: 2: 1 $

Young's modulus is given by $Y=\frac{F l}{A \Delta l}$ [where $l$ is length wire $\Delta l$ is elongation in wire]

$ Y_1=\frac{F_1 l_1}{A_1 \Delta l_1} \Rightarrow \Delta l_1=\frac{F_1 l_1}{A_1 Y_1} $

Similarly for $2^{\text {nd }}$ and $3^{\text {rd }}$ wire, elongation is

$ \begin{aligned} & \Delta l_2=\frac{F_2 l_2}{A_2 y_2}, \Delta l_3=\frac{F_3 l_3}{A_3 l_3} \\ & \Delta l_1: \Delta l_2: \Delta l_3=\frac{1}{3}: \frac{1}{4}: \frac{1}{3} \end{aligned} $

[Given, $F_1=F_2=F_3, l_1=l_2=l_3$ ]

Hence, ratio of elongation is

$ \Rightarrow \quad \Delta l_1: \Delta l_2: \Delta l_3:=4: 3: 4 $

The increase in length of the wire is

$ \begin{aligned} \delta L & =\frac{4 F L}{\pi D^2 y} \\ \Rightarrow \quad \delta L & =\left(\frac{L}{D^2}\right) k \end{aligned} $

Case $1 L=100 \mathrm{~cm}, D=1 \mathrm{~mm}$

$ \therefore \quad \delta L_1=\frac{100}{(1)^2} k=100 k $

Case $2 L=200 \mathrm{~cm}, D=2 \mathrm{~mm}$

$ \Delta L_2=\frac{200 k}{(2)^2}=50 k $

Case $3 L=300 \mathrm{~cm}, D=3 \mathrm{~mm}$

$ \Delta L_3=\frac{300 k}{(3)^2} k=33.3 k $

Case $4 L=50 \mathrm{~cm}$,

$ \begin{aligned} D & =0.5 \mathrm{~mm}=\frac{1}{2} \mathrm{~mm} \\ \Delta L_4 & =\frac{50}{\left(\frac{1}{2}\right)^2} k=200 k \end{aligned} $

So, among these option (d) is the correct option.

Using Young's modulus of elasticity we have,

$ \begin{aligned} Y & =\frac{F L}{A \Delta L} \\ \Rightarrow \quad \Delta L & =\frac{F L}{A Y} \end{aligned} $

( ∵ Since $\frac{F}{y}$ is same for same material)

$\therefore \Delta L \propto=\frac{L}{A}$

For wire, $A, \Delta L \propto \frac{1}{\pi\left(0.2 \times 10^{-3}\right)^2}$

$ \Rightarrow \quad \frac{1}{\pi(0.04)} \times 10^{-6} \mathrm{~m} $

For wire B,

$ \Delta L \propto \frac{2}{\pi\left(0.4 \times 10^{-3}\right)^2}=\frac{2}{\pi(0.16) \times 10^{-6}} \mathrm{~m} $

For wire C ,

$ \Delta L \propto \frac{3}{\pi\left(0.6 \times 10^{-3}\right)^2}=\frac{3}{\pi(0.36) \times 10^{-6}} \mathrm{~m} $

For wire D,

$ \Delta L \propto \frac{4}{\pi\left(0.1 \times 10^{-3}\right)^2}=\frac{4}{\pi(0.64) \times 10^{-6}} \mathrm{~m} $

It is clear that wire $A$ has maximum elongation ( $\Delta L$ ).

The ratio of the linear expansion coefficients of $A$ to $B$ is $\frac{3}{2}$.

The Young's modulus $Y$ is the same for both wires.

For a wire, the thermal expansion formula is:

$ \Delta l = l \alpha \Delta T $

where $\Delta l$ is the change in length, $l$ is the original length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature.

We use Young's modulus $Y$:

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

Stress is given by:

$ \text{Stress}_A = Y \times \alpha_A \Delta T $

For wire $A$, using the formula for Young’s modulus:

$ Y = \frac{(\text{Stress})_A}{l \times \alpha_A \times \Delta T} $

Similarly, for wire $B$:

$ Y = \frac{(\text{Stress})_B}{l \times \alpha_B \times \Delta T} $

Given that $Y$ is the same for both $A$ and $B$:

$ \frac{\text{Stress}_A}{\alpha_A \Delta T} = \frac{\text{Stress}_B}{\alpha_B \Delta T} $

Therefore, the ratio of the thermal stresses is determined by the ratio of their coefficients of linear expansion:

$ \frac{\text{Stress}_A}{\text{Stress}_B} = \frac{\alpha_A}{\alpha_B} = \frac{3}{2} $

Thus, the ratio of the thermal stresses produced in wires $A$ and $B$ is $3:2$.

To determine the strain on the wire, consider the following:

Original length of the wire: $ 40 \, \text{cm} $

Amount it is stretched (elongation): $ 0.1 \, \text{cm} $

The formula to calculate strain is:

$ \text{Strain} = \frac{\text{Elongation}}{\text{Original Length}} $

Substituting the given values:

$ \text{Strain} = \frac{0.1}{40} = 0.0025 $

To express this in scientific notation:

$ 0.0025 = 25 \times 10^{-4} $

Thus, the strain on the wire is $ 25 \times 10^{-4} $.

Work done in stretching a wire is

$ \begin{aligned} W & =\frac{1}{2} \times \text { stress } \times \text { strain × volume } \\ & =\frac{1}{2} \sigma \varepsilon A \cdot l=\frac{1}{2} Y \varepsilon^2 A \cdot l \\ \sigma & =\text { stress }=Y \cdot \varepsilon \\ \varepsilon & =\text { Strain } \frac{\Delta l}{l} \end{aligned} $

$A=$ Area of cross - section $=10^{-6} \mathrm{~m}^2$

$l=$ length of wire $=2 \mathrm{~m}$

$Y=$ Young's modulus $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

$\Delta l=$ change in length $=0.004 \mathrm{~m}$

So, work of stretching,

$ \begin{aligned} W & =\frac{1}{2} \times 2 \times 10^{11} \times \frac{\left(4 \times 10^{-3}\right)^2}{(2)^2} \times 10^{-6} \times 2 \\ & =8 \times 10^{-1}=0.8 \mathrm{~J} \end{aligned} $

Young's modulus of a rod is given by

$ Y=\frac{F \cdot l}{\Delta l \cdot A} $

where, $\Delta l=$ change in length.

$ \begin{array}{ll} \Rightarrow & \Delta l=\frac{F \cdot l}{Y \cdot A} \\ \text { or } & \Delta l=\frac{F \cdot l}{Y \pi r^2} \end{array} $

Substituting given values, we get

$ \begin{aligned} \Rightarrow \quad \Delta l & =\frac{10 \pi \times 10^3 \times 50 \times 10^{-2}}{2 \times 10^{11} \times \pi \times\left(10 \times 10^{-3}\right)^2} \\ & =25 \times 10^{4-2-11+4} \\ & =25 \times 10^{-5} \mathrm{~m} \\ & =25 \times 10^{-5} \times 10^3 \mathrm{~mm} \\ & =25 \times 10^{-2} \mathrm{~mm}=0.25 \mathrm{~mm} \end{aligned} $

Given,

For two wires,

$ \begin{aligned} & l_1=l_2=l, r_1=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\ & r_2=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \\ & \Delta l_2=2 \Delta l_1, F_1=F_2 \end{aligned} $

we know that,

$ \begin{aligned} & Y=\frac{F / A}{\Delta l / l}=\frac{F l}{A \Delta l} \\ & Y=\frac{F l}{\pi r^2 \Delta l} \end{aligned} $

$F$ and $l$ are same.

$ \begin{aligned} \therefore \quad Y & \propto \frac{1}{r^2 \Delta l} \\ \Rightarrow \quad \frac{Y_1}{Y_2} & =\frac{r_2^2 \Delta l_2}{r_1^2 \Delta l_1} \\ & =\frac{\left(1.5 \times 10^{-3}\right)^2 2 \Delta l_1}{\left(2 \times 10^{-3}\right)^2 \Delta l_1} \\ & =\frac{2 \times 2.25 \times 10^{-6}}{4 \times 10^{-6}} \\ & =\frac{2.25}{2}=\frac{9}{8} \end{aligned} $

Given, mass of object, $m=15 \mathrm{~kg}$

Length of wire, $l=1 \mathrm{~m}$

Area of cross-section of wire,

$ A=0.05 \mathrm{~cm}^2=5 \times 10^{-6} \mathrm{~m}^2 $

Angular velocity, $\omega=4 \mathrm{rad} / \mathrm{s}$

Young's modulus, $Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

At lowest point during the motion in vertical circle,

$ \begin{aligned} T-m g & =m \omega^2 l \\ T & =m g+m \omega^2 l \\ & =m\left(g+\omega^2 l\right) \\ & =15\left(10+4^2 \times l\right) \\ & =390 \mathrm{~N} \end{aligned} $

$ \begin{aligned} &\text { We know that, Young's modulus, }\\ &\begin{aligned} Y & =\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\Delta l / l} \\ \Rightarrow \quad Y & =\frac{T l}{A \Delta l} \\ \Rightarrow \quad \Delta l & =\frac{T l}{A Y} \\ & =\frac{390 \times 1}{5 \times 10^{-6} \times 2 \times 10^{11}} \\ & =390 \times 10^{-6} \mathrm{~m} \\ & =0.39 \times 10^{-3} \mathrm{~m} \\ & =0.39 \mathrm{~mm} \end{aligned} \end{aligned} $

Pressure on the bottom of swimming pool

$ p=\rho g h=1000 \times 10 \times 22 $

Now, we know that, the bulk modulus is given by

$ \begin{gathered} B=\frac{p}{\left(\frac{\Delta V}{V}\right)} \\ \Rightarrow \frac{\Delta V}{V}=\frac{p}{B}=\frac{1000 \times 10 \times 22}{2.2 \times 10^9} \\ \Rightarrow \frac{\Delta V}{V}=100000 \times 10^{-9} \Rightarrow \frac{\Delta V}{V}=10^{-4} \end{gathered} $

By definitions of shear modulus, it matches with resistance to change against deformation force.

shearing stress is the tangential stress applied on any surface.

Elastic fatigue itself means temporary loss of elastic property.

Modulus of elasticity is the proportionality constant in Hooke's law.

If a rod or wire of length $L$ and cross-section area $A$ under the action of stretching force $F$ applied normally to its face suffers a increase $\Delta L$ in length, then

$ Y=\frac{\text { tensile stress }}{\text { longitudinal strain }}=\frac{F / A}{\frac{\Delta L}{L}} $

Hence, Young's modulus is proportionality constant that relates

the force per unit area applied perpendicularly at the surface of an object to the fractional change in length.

For metal wire $A, L_A=L, R_A=R$

For metal wire $B, L_B=3 L, R_B=2 R$ When the ends of the of metal wire $A$ and $B$ are joined and pulled with a constant force $F$ after fixing the one end of combined system, then change in length of wire $A$ and $B$ is same.

$ \begin{array}{ll} \text { i.e., } & \Delta L_A=\Delta L_B \\ \Rightarrow & \frac{F L_A}{A_A Y_A}=\frac{F L_B}{A_B Y_B} \\ \Rightarrow & \frac{F L}{\pi R_A^2 Y_1}=\frac{F \times 3 L}{\pi R_B^2 Y_2} \\ \Rightarrow & \frac{1}{\pi R^2 Y_A}=\frac{3}{\pi(2 R)^2 Y_B} \\ \Rightarrow & \frac{1}{\pi R^2 Y_A}=\frac{3}{4 \pi R^2 Y_B} \\ \Rightarrow & \frac{1}{Y_A}=\frac{3}{4 Y_B} \Rightarrow \frac{Y_B}{Y_A}=\frac{3}{4} \end{array} $

As $Y=$ Young's modulus of material

$ \begin{aligned} & \qquad=\frac{F / A}{\Delta L / L_0} \\ & \Rightarrow \quad \Delta L=\frac{L_0 F}{Y A} \Rightarrow \quad L-L_0=\frac{L_0 F}{Y A} \\ & \Rightarrow \quad L=L_0+\frac{L_0 F}{Y A} \\ & \text { Here, } L_1=L_0+\frac{L_0 T_1}{Y A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \quad L_2=L_0+\frac{L_0 T_2}{Y A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & \left(L_1-L_0\right) T_2=\frac{L_0 T_1 T_2}{Y A} \text { and } \\ & \left(L_2-L_0\right) T_1=\frac{L_0 T_1 T_2}{Y A} \end{aligned} $

So, equating we have

$ \begin{array}{lr} \Rightarrow & \left(L_1-L_0\right) T_2=\left(L_2-L_0\right) T_1 \\ \Rightarrow & L_1 T_2-L_2 T_1=L_0\left(T_2-T_1\right) \\ \Rightarrow & L_0=\frac{L_1 T_2-L_2 T_1}{T_2-T_1} \end{array} $

The given situation is shown in the following figure.

$ \begin{aligned} & \text { Shear strain }=\frac{x}{l}=\frac{0.2 \mathrm{~mm}}{50 \mathrm{~cm}}=\frac{0.2 \times 10^{-3}}{50 \times 10^{-2}} \\ & \text { Shear stress }=\frac{F}{A}=\frac{10^5}{\frac{50}{100} \times \frac{10}{100}} \\ & =\frac{10^5 \times 10^4}{5 \times 10^2}=2 \times 10^6 \\ & \text { Modulus of rigidity }=\frac{\text { Stress }}{\text { Strain }} \\ & =\frac{2 \times 10^6 \times 50 \times 10^{-2}}{0.2 \times 10^{-3}} \\ & =5 \times 10^9 \mathrm{~N} / \mathrm{m}^2=5 \mathrm{GPa} \end{aligned} $