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} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} Y & =\frac{W l}{A \Delta l}=\left(\frac{W}{\Delta l}\right) \frac{l}{A} \\ & =\frac{(20-0)}{(1-0) \times 10^{-3}} \times \frac{1}{1 \times 10^{-6}} \\ & =2 \times 10^{10} \mathrm{~N} / \mathrm{m}^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & Y=\frac{\text { stress }}{\text { strain }}=\text { slope of stress-strain curve } \\ & =\tan \theta \\ & \therefore \frac{Y_A}{Y_B}=\frac{\tan \theta_1}{\tan \theta_2} \\ & =\frac{\tan 30^{\circ}}{\tan 60^{\circ}}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{\sqrt{3} \times \sqrt{3}} \\ & \Rightarrow \frac{Y_A}{Y_B}=\frac{1}{3} \\ & \Rightarrow Y_B=3 Y_A \end{aligned} $

$ \begin{aligned} &\text { } U=\frac{1}{2} \times \text { stress } \text { × } \text { strain } \text { × } \text { volume }\\ &\begin{aligned} & =\frac{1}{2} \times\left(Y \times \frac{\Delta l}{l}\right) \times\left(\frac{\Delta l}{l}\right) \times(A \times l) \\ & =\frac{1}{2} \times Y \times A \times \frac{\Delta l^2}{l} \\ & =\frac{1}{2} \times 1.2 \times 10^{11} \times \frac{1 \times 10^{-6} \times 1 \times 10^{-6}}{1} \\ & =6 \times 10^{-2} \mathrm{~J} \end{aligned} \end{aligned} $

As contraction in wire is prevented by extending it with a load,

Thermal contraction $=$ Extension due to load

$ \begin{array}{ll} \Rightarrow & \alpha l \Delta T=\frac{F l}{Y A} \Rightarrow \alpha l \Delta T=\frac{m g l}{Y A} \\ \Rightarrow & m=\frac{\alpha \Delta T \cdot Y A}{g}=\frac{\alpha \Delta T Y A}{g} \end{array} $

Substituting given values,

$ \begin{aligned} m & =\frac{10^{-5} \times 100 \times 10^{11} \times 4 \times 10^{-6}}{10} \\ & =4 \times 10=40 \mathrm{~kg} \end{aligned} $

Step 1: Formula for Elongation

The amount a wire stretches depends on Young's modulus formula:

$ Y=\frac{F l}{\Delta l A}=\frac{4 F l}{\Delta l \pi d^2} $

Step 2: What stays the same?

Both wires are made of the same material, have the same length, and the same stretching force. This means their Young's modulus ($Y$), original length ($l$), and force ($F$) are the same.

Step 3: Setting up the equation for both wires

Since Young's modulus is the same for both wires, we set the formulas equal for each wire:

$ \begin{aligned} Y_1 = Y_2 \\ \frac{4 F l}{\Delta l_1 \pi d_1^2} = \frac{4 F l}{\Delta l_2 \pi d_2^2} \\ \end{aligned} $

Step 4: Rearranging to compare elongations

We can simplify to find how the elongation ($\Delta l$) relates to diameter ($d$):

$ \Delta l_1 \cdot d_1^2 = \Delta l_2 \cdot d_2^2 $

So,

$ \Delta l_2 = \Delta l_1 \cdot \left(\frac{d_1}{d_2}\right)^2 $

Step 5: Substitute values

We are told $\Delta l_1 = 1$ mm (the first wire stretches 1 mm) and $d_2 = 4 d_1$ (the second wire's diameter is 4 times bigger).

$ \Delta l_2 = 1 \cdot \left(\frac{d_1}{4d_1}\right)^2 = 1 \cdot \left(\frac{1}{4}\right)^2 = \frac{1}{16}~\text{mm} $

Final Answer

The second, thicker wire stretches only $\frac{1}{16}$ mm when the same force is applied.

• Longitudinal strain, $ \varepsilon_L = 0.2\% = 0.002 $

• Poisson’s ratio, $ \mu = 0.3 $

We are asked to find the volume strain ($ \varepsilon_V $).

Step 1: Recall the formula for volume strain

For small deformations,

$ \text{Volume strain} = \varepsilon_V = \varepsilon_L (1 - 2\mu) $

However, to be more precise, the change in volume $ \Delta V / V = \varepsilon_L + 2\varepsilon_t $,

where $ \varepsilon_t = -\mu \varepsilon_L $ (lateral strain is negative if the material stretches longitudinally).

So,

$ \frac{\Delta V}{V} = \varepsilon_L + 2(-\mu \varepsilon_L) $

$ \Rightarrow \varepsilon_V = \varepsilon_L (1 - 2\mu) $

Step 2: Substitute the values

$ \varepsilon_V = 0.002 (1 - 2 \times 0.3) $

$ \varepsilon_V = 0.002 (1 - 0.6) $

$ \varepsilon_V = 0.002 \times 0.4 = 0.0008 $

Step 3: Convert to percentage

$ \varepsilon_V = 0.0008 \times 100 = 0.08\% $

✅ Final Answer:

$ \boxed{0.08\%} $

Correct Option: (B) 0.08%

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \text { Compressibility }=\frac{1}{\text { Bulk modulus }(B)} \\ & =\frac{-\frac{1}{\Delta P}}{\frac{\Delta V}{V}}=-\frac{\frac{\Delta V}{V}}{\Delta P} \\ & =\frac{-(-0.0025)}{(250-200) \times 10^3}=5 \times 10^{-8} \end{aligned} \end{aligned} $

Elastic potential energy per unit volume

$ \begin{aligned} U & =\frac{W}{V}=\frac{\frac{1}{2} F \Delta l}{A \cdot l} \\ & =\frac{1}{2} \cdot \frac{F}{A} \cdot \frac{\Delta l}{l} \\ & =\frac{1}{2} \times \text { Stress } \times \text { Strain } \\ & =\frac{1}{2} \times \text { Stress } \times \frac{\text { Stress }}{Y} \\ & =\frac{1}{2} \times S \times \frac{S}{Y}=\frac{S^2}{2 Y} \end{aligned} $

$ \text { } \begin{aligned} Y & =\frac{F l}{A \Delta l} \\ F & =\frac{Y A \Delta l}{l}=\frac{Y A(2 l-l)}{l} \\ & =A Y=1 \times 10^{-6} \times 2 \times 10^{11} \\ & =2 \times 10^5 \mathrm{~N} \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & Y=\frac{F L}{\pi r^2 \Delta L} \\ \Rightarrow & \Delta L=\frac{F L}{\pi r^2 Y} \\ \Rightarrow & \Delta L \propto \frac{F L}{r^2} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \frac{\Delta L_2}{\Delta L_1} & =\frac{F_2}{F_1} \times \frac{L_2}{L_1} \times\left(\frac{r_1}{r_2}\right)^2 \\ & =\left(\frac{F_{1 / 2}}{F_1}\right) \times\left(\frac{L_1}{L}\right)\left(\frac{r_1}{r_{1 / 2}}\right)^2 \end{aligned} $

$ \begin{array}{rlr} \Rightarrow \quad \frac{\Delta L_2}{\Delta L_1} & =\frac{1}{2} \times 1 \times 2^2 & \\ \Delta L_2 & =2 \Delta L_1 & {\left[\because \Delta L_1=L\right]} \end{array} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =2 L $

• Depth of sea, $ h = 1\,\text{km} = 1000\,\text{m} $

• Relative decrease in volume, $ \frac{\Delta V}{V} = 0.01\% = 0.0001 = 10^{-4} $

• Acceleration due to gravity, $ g = 10\,\text{m/s}^2 $

• Density of sea water, $ \rho \approx 10^3\,\text{kg/m}^3 $

Step 1. Calculate the pressure at that depth

$ P = \rho g h = 1000 \times 10 \times 1000 = 10^7\,\text{N/m}^2 $

Step 2. Use the definition of Bulk modulus $ K $

$ K = - \frac{\Delta P}{\Delta V / V} $

Here, $ \Delta P = P = 10^7\,\text{N/m}^2 $, and $ \frac{\Delta V}{V} = -10^{-4} $

$ K = \frac{10^7}{10^{-4}} = 10^{11}\,\text{N/m}^2 $

✅ Final Answer:

$ \boxed{K = 10^{11}\,\text{N/m}^2} $

Correct Option: D) $ 10 \times 10^{10} \,\text{N/m}^2 $

Elongations in $A$ and $B$ are same $\Rightarrow$

$ \begin{aligned} & & \Delta l_A & =\Delta l_B \\ & & \frac{T_A l_A}{A_A \cdot Y_A} & =\frac{T_B l_B}{A_B \cdot Y_B} \\ & \Rightarrow & \frac{T_A}{r_A^2 Y_A} & =\frac{T_B}{r_B^2 Y_B} \\ & \Rightarrow & \frac{T_A}{T_B} & =\left(\frac{r_A}{r_B}\right)^2-\left(\frac{Y_A}{Y_B}\right) \end{aligned} $

Given, $\frac{r_A}{r_B}=\frac{2}{3}$ and $\frac{Y_A}{Y_B}=\frac{3}{2}$

$ \therefore \quad \frac{T_A}{T_B}=\left(\frac{2}{3}\right)^2\left(\frac{3}{2}\right)=\frac{2}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For rotational equilibrium of bar; taking moments about point of suspension ' $P$ ',

$ \begin{aligned} & T_A \cdot x=T_B(l-x) \\ & \Rightarrow \quad \frac{l-x}{x}=\frac{T_A}{T_B}=\frac{2}{3} \,\,\,\,[form Eq. (i)]\\ & \Rightarrow \quad \frac{150-x}{x}=\frac{2}{3} \\ & \quad 150 \times 3=5 x \,\,\,\,\,\,\, [\therefore l=150 \mathrm{~cm}]\end{aligned} $

or

$ x=90 \mathrm{~cm}(\text { from } P) $

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 .

To find the strain produced in the wire, follow these calculations:

Volume Conversion:

Given volume $ V = 2 \, \text{cm}^3 $.

Convert this to cubic meters:

$ V = 2 \times 10^{-6} \, \text{m}^3 $

Energy Density Calculation:

The work done on the wire is given as $ W = 16 \times 10^2 \, \text{J} $.

Calculate the energy density $ u $:

$ u = \frac{W}{V} = \frac{16 \times 10^2}{2 \times 10^{-6}} = 8 \times 10^8 \, \text{J/m}^3 $

Strain Calculation:

Using the formula for energy density:

$ u = \frac{1}{2} \times Y \times (\text{strain})^2 $

where $ Y = 4 \times 10^{12} \, \text{N/m}^2 $ is Young's modulus.

Rearrange to solve for strain:

$ \text{strain} = \sqrt{\frac{2u}{Y}} = \sqrt{\frac{2 \times 8 \times 10^8}{4 \times 10^{12}}} $

Simplify the expression:

$ \text{strain} = 2 \times 10^{-2} = 0.02 $

Therefore, the strain produced in the wire is $ 0.02 $.

Given:

Length of the wire (L): 100 cm = 1 m

Cross-sectional area (A):

$ A = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2 $

Force (F): 440 N

Elongation of the wire ($\Delta L$):

$ \Delta L = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} $

Calculation of Young's Modulus:

Young's modulus $Y$ is given by the formula:

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

Substitute the given values:

$ Y = \frac{F \cdot L}{A \cdot \Delta L} = \frac{440 \times 1}{2 \times 10^{-6} \times 2 \times 10^{-3}} $

Simplify:

$ Y = \frac{440}{4} \times 10^9 = 110 \times 10^9 $

Hence, the Young's modulus $Y$ is:

$ Y = 1.1 \times 10^{11} \, \text{N/m}^2 $

Given the problem, we are to determine the elongation of a copper wire with the following specifications:

Cross-sectional area: $3.5 \, \text{mm}^2$

Mass of block A ($m_A$): $4 \, \text{kg}$

Mass of block C ($m_C$): $7 \, \text{kg}$

Young's Modulus for Copper, $Y$: $10 \times 10^{10} \, \text{N/m}^2$

Acceleration due to gravity, $g$: $10 \, \text{m/s}^2$

To calculate the elongation ($\Delta L$) of the copper wire, use the formula for linear elongation induced by a force:

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

Here, the force $F$ is due to the weight of block C, which is calculated as $F = m_C \cdot g$. The length $L$ is given as $0.5 \, \text{m}$.

Substituting the given values:

$ \Delta L = \frac{7 \cdot 9.8 \cdot 0.5}{10^{11} \cdot 3.5 \times 10^{-6}} $

Calculate the result:

$ \Delta L = 10^{-4} \, \text{m} $

Thus, the elongation of the copper wire is $10^{-4} \, \text{m}$.

To calculate the strain in the wire, let's begin with the given data:

Mass, $ m = 4 \, \text{kg} $

Velocity, $ v = 12 \, \text{m/s} $

Length of wire = $ 4 \, \text{m} $

Diameter = $ 2.0 \, \text{mm} $ or radius = $ 1.0 \times 10^{-3} \, \text{m} $

Young's modulus of steel = $ 2 \times 10^{11} \, \text{N/m}^2 $

Step-by-Step Calculation:

Calculate the Cross-Sectional Area of the Wire:

$ \text{Area} = \pi r^2 = \pi \times (1.0 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 $

Calculate the Centripetal Force required to revolve the stone in a horizontal circle:

$ F_c = \frac{m v^2}{r} = \frac{4 \times 12^2}{4} = 144 \, \text{N} $

(Note: The original calculation contained an error with the radius used in centripetal force.)

Calculate the Stress on the Wire:

$ \text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{144}{\pi \times 10^{-6}} = \frac{144}{3.14159 \times 10^{-6}} \, \text{Nm}^{-2} $

Simplifying:

$ \text{Stress} \approx 4.59 \times 10^7 \, \text{Nm}^{-2} $

Calculate the Strain using Young's Modulus:

$ \text{Strain} = \frac{\text{Stress}}{\text{Young's modulus}} = \frac{4.59 \times 10^7}{2 \times 10^{11}} $

Solving for strain:

$ \text{Strain} = 2.295 \times 10^{-4} $

Rounding off to one decimal place gives:

$ \text{Strain} \approx 2.4 \times 10^{-4} $

The strain in the wire is approximately $ 2.4 \times 10^{-4} $.

Given the following:

Longitudinal strain $ \varepsilon $

Young's modulus $ Y $

The relationship between stress and Young's modulus is given by:

$ \text{Young's modulus} = \frac{\text{stress}}{\text{strain}} $

From this, we can express stress as:

$ \text{Stress} = Y \cdot \varepsilon $

The formula for elastic energy stored per unit volume is:

$ U = \frac{1}{2} \times \text{stress} \times \text{strain} $

Substituting in the expression for stress, we get:

$ U = \frac{1}{2} \times Y \cdot \varepsilon \times \varepsilon = \frac{Y \varepsilon^2}{2} $

When the load on a wire is increased from 5 kg to 8 kg, the elongation of the wire changes from 1 mm to 1.8 mm. To find the work done during this elongation, we use the acceleration due to gravity as $10 \, \mathrm{ms}^{-2}$.

First, calculate the work done for the initial elongation:

The force due to a 5 kg load is:

$ 5 \, \mathrm{kg} \times 9.8 \, \mathrm{N/kg} = 49 \, \mathrm{N} $

Work done for an elongation of 1 mm ($0.001 \, \mathrm{m}$) under this force is:

$ W_1 = \frac{1}{2} \times 49 \times 0.001 = 0.0245 \, \mathrm{J} = 24.5 \times 10^{-3} \, \mathrm{J} $

Now, calculate the work done for the increased elongation:

The force due to an 8 kg load is:

$ 8 \, \mathrm{kg} \times 9.8 \, \mathrm{N/kg} = 78.4 \, \mathrm{N} $

Work done for an elongation of 1.8 mm ($0.0018 \, \mathrm{m}$) under this force is:

$ W_2 = \frac{1}{2} \times 78.4 \times 0.0018 = 0.07056 \, \mathrm{J} = 70.56 \times 10^{-3} \, \mathrm{J} $

The work done during the additional elongation (from 1 mm to 1.8 mm) is the difference between these two energies:

$ \Delta W = W_2 - W_1 = 70.56 \times 10^{-3} - 24.5 \times 10^{-3} = 46.06 \times 10^{-3} \, \mathrm{J} $

This value rounds to $47 \times 10^{-3} \, \mathrm{J}$.

The work done in stretching a wire can be calculated by considering the relation:

$ W = \frac{1}{2}\left(\frac{F}{A}\right) \text{ (strain) (volume) } $

Breaking it down further:

$ W = \frac{1}{2}\left(\frac{F}{A}\right)\left(\frac{e}{l}\right) \times A \times l $

This simplifies to:

$ W = \frac{1}{2} F e $

We deduce that:

$ W \propto F \propto \frac{V A e}{l} $

Since:

$ W \propto \frac{r^2}{l} $

Let's compare the initial and new wire conditions:

For the first wire, we have the following proportion:

$ \frac{2}{W_2} = \frac{r_1^2}{r_2^2} \times \frac{l_2}{l_1} $

Plugging in the values given:

$ = \frac{r^2}{4r^2} \times \frac{l}{2l} = \frac{1}{8} $

Solving for $ W_2 $:

$ W_2 = 8 \times 2 = 16 \, \text{J} $

Hence, the work necessary to stretch the second wire is 16 Joules.

To solve this problem, we start by examining the given relationships for the two copper wires, A and B.

Given Ratios:

Lengths: $ \frac{l_A}{l_B} = \frac{1}{2} $

Diameters: $ \frac{d_A}{d_B} = \frac{3}{2} $ implies $ \frac{r_A}{r_B} = \frac{3}{2} $. Thus, $ \frac{r_A^2}{r_B^2} = \frac{9}{4} $.

Forces: $ \frac{F_A}{F_B} = \frac{3}{1} $

Elastic Potential Energy Per Unit Volume:

The elastic potential energy per unit volume in the wire is given by:

$ U = \frac{1}{2} \frac{F \cdot \Delta L}{V} $

Young's Modulus Definition:

From Young's modulus $ Y $, we have:

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

Volume:

$ V = A \cdot L $

Substituting $\Delta L$ and $V$ into the expression for $U$:

$ U = \frac{1}{2} \cdot \frac{F \cdot F \cdot L}{A \cdot L \cdot A \cdot Y} = \frac{1}{2} \frac{F^2}{A^2 Y} $

Ratio of Elastic Potential Energies $ U_A $ and $ U_B $:

$ \frac{U_A}{U_B} = \frac{\frac{1}{2} \frac{F_A^2}{(\pi r_A^2)^2 Y}}{\frac{1}{2} \frac{F_B^2}{(\pi r_B^2)^2 Y}} = \frac{F_A^2 (r_B^2)^2}{F_B^2 (r_A^2)^2} $

Now plug in the ratios:

$ \frac{U_A}{U_B} = \left(\frac{3}{1}\right)^2 \cdot \left(\frac{4}{9}\right)^2 $

Calculate the final ratio:

$ \frac{U_A}{U_B} = \frac{9}{1} \cdot \frac{16}{81} $

$ \frac{U_A}{U_B} = \frac{144}{81} $

$ \frac{U_A}{U_B} = \frac{16}{9} $

Thus, the ratio of elastic potential energies stored per unit volume in wires A and B is $ 16:9 $.

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.

Two wires, A and B, with the same cross-sectional area, are connected end-to-end. When the same tension is applied to both wires, wire B elongates twice as much as wire A. Let the initial lengths of wires A and B be $ L_A $ and $ L_B $ respectively. The Young's modulus for wire A is $ 2 \times 10^{11} \ \mathrm{N/m^2} $ and for wire B is $ 1.1 \times 10^{11} \ \mathrm{N/m^2} $.

Let's illustrate this situation with the following image:

We are given that the elongation in wire B is twice that of wire A when the same tension $ T $ is applied. This can be expressed mathematically as:

$ \Delta L_B = 2 \Delta L_A $

The Young's modulus formulas for wires A and B are as follows:

$ Y_A = \frac{T L_A}{A \cdot \Delta L_A} \quad \text{and} \quad Y_B = \frac{T L_B}{A \cdot \Delta L_B} $

By taking the ratio of these two equations, we obtain:

$ \frac{Y_A}{Y_B} = \frac{L_A}{L_B} \times \frac{\Delta L_B}{\Delta L_A} $

Substitute the given values:

$ \frac{2 \times 10^{11}}{1.1 \times 10^{11}} = \frac{L_A}{L_B} \times \frac{2 \Delta L_A}{\Delta L_A} $

This simplifies to:

$ \frac{2}{1.1} = \frac{L_A}{L_B} \times 2 $

Therefore:

$ \frac{L_A}{L_B} = \frac{2}{1.1 \times 2} = \frac{1}{1.1} \approx \frac{10}{11} $

Thus, the ratio of the initial lengths of wires A and B is approximately $ \frac{L_A}{L_B} = \frac{10}{11} $.

We know that, young's modulus,

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

Where, $A=$ Area of cross-section, $l=$ length of wire and $\Delta l=$ elongation in the longest wire.

Here, $A=\pi r^2$

$=\pi\left(\frac{d}{2}\right)^2=\pi \frac{d^2}{4} \quad(\because d=2 r)$

$\begin{aligned} & \therefore \quad Y=\frac{F l}{\frac{\pi d^2}{4} \Delta l} \\ & \Rightarrow \Delta l=\frac{4 F l}{\pi d^2 Y}=\left(4 \frac{F}{\pi Y}\right) \frac{l}{d^2} \Rightarrow \Delta l=k \frac{l}{d^2} \end{aligned}$

where, $k=\frac{4 F}{\pi y}=$ constant

For wire given in option (a)

$(\Delta)_1=k \cdot \frac{50}{(0.05)^2}=20000 \mathrm{k}$

similarly,

For wire given in option (b)

$(\Delta)_2=k \times \frac{200}{(0.2)^2}=5000 \mathrm{k}$

For wire given in option (c)

$\Delta l_3=k \cdot \frac{300}{(0.3)^2}=3333.33 \mathrm{~k}$

For wire given in option (d)

$\Delta l_4=\frac{100}{(0.1)^2}=10000 \mathrm{~k}$

Hence, we see that wire given in option (a) has maximum elongation.

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

Applied force, $F=2 \times 10^{11} \mathrm{~N}$

Cross-sectional area, $A=1 \mathrm{~m}^2$

Let initial and final length be $L_1=L$ and $L_2=$ ?

$\begin{array}{ll} \text { Since, } Y= & \frac{F L_1}{A\left(L_2-L_1\right)} \\ \therefore & \frac{L_2-L_1}{L_1}=\frac{2 \times 10^{11}}{2 \times 10^{11}}=1 \\ \Rightarrow & \frac{L_2}{L_1}-1=1 \Rightarrow \frac{L_2}{L_1}=2 \\ \Rightarrow & L_2=2 L_1 \\ \therefore & L_2=2 L \end{array}$

Given, length of rubber,

$\begin{aligned} l & =12 \mathrm{~cm} \\ & =12 \times 10^{-2} \mathrm{~m} \end{aligned}$

Density of rubber, $\rho=1.5 \mathrm{~kg} \mathrm{~m}^{-3}$

Young's modulus, $Y=5 \times 10^8 \mathrm{~N} / \mathrm{m}^2$

As we know that,

$Y=\frac{m g l}{A \Delta l} \times \frac{1}{2}$

Centre of mass of rod is at the centre of rod

$\therefore \quad Y=\frac{\rho(A \cdot l) \cdot g \cdot l}{2 A \cdot \Delta l}$

where, $A$ is area of cross-section and $g$ is acceleration due to gravity.

$\begin{aligned} \therefore \quad Y & =\frac{\rho l^2 g}{2 \Delta l} \Rightarrow \Delta l=\frac{\rho l^2 g}{2 Y} \\ & =\frac{1.5 \times\left(12 \times 10^{-2}\right)^2 \times 10}{2 \times 5 \times 10^8}=2.16 \times 10^{-10} \mathrm{~m}\end{aligned}$

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} $