Elasticity

$ \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) $

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 $.

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.

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