Young’s Modulus Questions in English

(B) According to Hooke's law,the relationship between stress and strain is given by the modulus of elasticity.
Stress is defined as the force per unit area,having dimensions $[M L^{-1} T^{-2}]$.
Strain is defined as the ratio of change in dimension to the original dimension,which is a dimensionless quantity.
The modulus of elasticity is defined as the ratio of stress to strain:
$\text{Modulus of Elasticity} = \frac{\text{Stress}}{\text{Strain}}$
Since strain is dimensionless,the dimensions of the modulus of elasticity are the same as the dimensions of stress.
Therefore,the modulus of elasticity is dimensionally equivalent to stress.

(C) The formula for the change in length (elongation) $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$.
Given:
Length $L = 1 \, m$
Area $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$
Mass $M = 1 \, kg$,so Force $F = Mg = 1 \times 10 = 10 \, N$
Young's Modulus $Y = 2 \times 10^{11} \, N/m^2$
Substituting the values:
$\Delta L = \frac{10 \times 1}{(1 \times 10^{-6}) \times (2 \times 10^{11})}$
$\Delta L = \frac{10}{2 \times 10^5} = 5 \times 10^{-5} \, m$
Converting to $mm$:
$\Delta L = 5 \times 10^{-5} \times 10^3 \, mm = 0.05 \, mm$.

(C) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,we have $l \propto \frac{F}{r^2}$.
Given the initial state: $l_1 = 1 \ mm$,$F_1 = W$,and $r_1 = r$.
Given the final state: $F_2 = 4W$ and $r_2 = 2r$.
Using the proportionality $l \propto \frac{F}{r^2}$,we get the ratio:
$\frac{l_2}{l_1} = \frac{F_2}{F_1} \times \left( \frac{r_1}{r_2} \right)^2$
Substituting the values: $\frac{l_2}{1} = \frac{4W}{W} \times \left( \frac{r}{2r} \right)^2 = 4 \times \left( \frac{1}{2} \right)^2 = 4 \times \frac{1}{4} = 1$.
Therefore,the new extension $l_2 = 1 \ mm$.

(C) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain.
$Y = \frac{\text{Stress}}{\text{Strain}}$
Since strain is a dimensionless quantity (ratio of change in length to original length),the unit of Young's modulus is the same as the unit of stress.
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
The $SI$ unit of force is Newton $(N)$ and the $SI$ unit of area is square meter $(m^2)$.
Therefore,the unit of Young's modulus is $N/m^2$ or $N m^{-2}$.

(B) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since the wires are similar ($L$ is same) and are under the same load ($F$ is same) and made of the same material ($Y$ is same),we have $l \propto \frac{1}{A}$.
Therefore,$\frac{A_2}{A_1} = \frac{l_1}{l_2}$.
Given $l_1 = 0.1 \ mm$,$l_2 = 0.05 \ mm$,and $A_1 = 4 \ mm^2$.
Substituting these values: $A_2 = A_1 \times \left( \frac{l_1}{l_2} \right) = 4 \times \left( \frac{0.1}{0.05} \right) = 4 \times 2 = 8 \ mm^2$.

(C) The elongation $\Delta L$ in a wire is given by the formula: $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since $F$,$L$,and $\Delta L$ are constant for both wires,we have $r^2 Y = \text{constant}$,which implies $r^2 \propto \frac{1}{Y}$.
Therefore,$\frac{r_2}{r_1} = \sqrt{\frac{Y_1}{Y_2}}$.
Given $Y_1 = 7 \times 10^{10}\, N/m^2$,$Y_2 = 12 \times 10^{10}\, N/m^2$,and diameter $d_1 = 3\, mm$ (so $r_1 = 1.5\, mm$).
$\frac{r_2}{1.5} = \sqrt{\frac{7 \times 10^{10}}{12 \times 10^{10}}} = \sqrt{\frac{7}{12}} \approx 0.7637$.
$r_2 = 1.5 \times 0.7637 \approx 1.145\, mm$.
The diameter $d_2 = 2 \times r_2 = 2 \times 1.145 = 2.29\, mm$.

(C) Given:
Diameter $d = 0.6 \, mm = 0.6 \times 10^{-3} \, m$,so radius $r = 0.3 \times 10^{-3} \, m$.
Young's modulus $Y = 0.9 \times 10^{11} \, N/m^2$.
Strain $\frac{\Delta L}{L} = 0.2\% = \frac{0.2}{100} = 0.002$.
Area of cross-section $A = \pi r^2 = \pi \times (0.3 \times 10^{-3})^2 = \pi \times 0.09 \times 10^{-6} \, m^2$.
Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$,we get $F = Y \times A \times \frac{\Delta L}{L}$.
Substituting the values: $F = (0.9 \times 10^{11}) \times (\pi \times 0.09 \times 10^{-6}) \times (0.002)$.
$F = 0.9 \times 10^{11} \times 3.1416 \times 0.09 \times 10^{-6} \times 0.002$.
$F \approx 0.9 \times 0.09 \times 3.1416 \times 200 \approx 50.89 \, N$.
Thus,the force required is nearly $51 \, N$.

(B) Young's modulus $(Y)$ is defined as the ratio of stress to strain.
$Y = \frac{\text{stress}}{\text{strain}}$
Let the original length of the wire be $L$. Since the length is doubled,the new length becomes $2L$.
Change in length $\Delta L = 2L - L = L$.
Strain is defined as the ratio of change in length to the original length:
$\text{strain} = \frac{\Delta L}{L} = \frac{L}{L} = 1$.
Given stress = $20 \times 10^8 \ N/m^2$.
Substituting these values into the formula for Young's modulus:
$Y = \frac{20 \times 10^8 \ N/m^2}{1} = 20 \times 10^8 \ N/m^2$.
Therefore,the correct option is $B$.

(D) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{\pi r^2 Y}$,where $F$ is the applied force,$L$ is the original length,$r$ is the radius,and $Y$ is Young's modulus.
Since $F$,$L$,and $Y$ remain constant,we have $l \propto \frac{1}{r^2}$.
Given the initial radius $r_1 = r$ and initial extension $l_1 = 2.00 \, mm$.
The new radius is $r_2 = r/2$.
Using the ratio: $\frac{l_2}{l_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r}{r/2} \right)^2 = (2)^2 = 4$.
Therefore,the new extension $l_2 = 4 \times l_1 = 4 \times 2.00 \, mm = 8.00 \, mm$.

(B) The force constant $K$ between atoms is related to Young's modulus $Y$ and interatomic spacing $r_0$ by the relation $K = Y \times r_0$.

Given, $Y = 20 \times 10^{10} \text{ N/m}^2$ and $r_0 = 3.0 \mathring{A} = 3.0 \times 10^{-10} \text{ m}$.
Substituting the values:
$K = (20 \times 10^{10} \text{ N/m}^2) \times (3.0 \times 10^{-10} \text{ m})$
$K = 60 \text{ N/m}$.
To express this in $N/\mathring{A}$:
$K = 60 \text{ N/m} = 60 \text{ N} / (10^{10} \mathring{A})$
$K = 6 \times 10^1 \times 10^{-10} N/\mathring{A}$
$K = 6 \times 10^{-9} N/\mathring{A}$.

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the change in length.
Rearranging for $\Delta L$: $\Delta L = \frac{F \cdot L}{A \cdot Y}$.
Given values:
$F = 4.8 \times 10^3 \, N$
$L = 4.0 \, m$
$A = 1.2 \, cm^2 = 1.2 \times 10^{-4} \, m^2$
$Y = 1.2 \times 10^{11} \, N/m^2$
Substituting the values:
$\Delta L = \frac{4.8 \times 10^3 \times 4.0}{(1.2 \times 10^{-4}) \times (1.2 \times 10^{11})}$
$\Delta L = \frac{19.2 \times 10^3}{1.44 \times 10^7}$
$\Delta L = 13.33 \times 10^{-4} \, m = 1.33 \times 10^{-3} \, m = 1.33 \, mm$.

(B) When the temperature of the rod increases by $\Delta t$,it tends to expand by an amount $\Delta L = L \alpha \Delta t$.
Since the rod is clamped between two rigid supports,this expansion is prevented,resulting in a compressive strain $e = \frac{\Delta L}{L} = \alpha \Delta t$.
According to Hooke's Law,the stress $\sigma$ is given by $\sigma = Y e$,where $Y$ is the Young's modulus.
Substituting the strain,we get $\sigma = Y \alpha \Delta t$.
Since stress is defined as force per unit area $(\sigma = \frac{F}{A})$,the force $F$ exerted on the supports is $F = \sigma A$.
Therefore,$F = Y A \alpha \Delta t$.

(A) The Young's modulus $Y$ is defined as the ratio of longitudinal stress to longitudinal strain.
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{Strain}}$
Rearranging the formula to solve for strain:
$\text{Strain} = \frac{F}{AY}$
Thus,the compressive strain on the plank is $F/AY$.

(D) The Young's modulus $(Y)$ can be calculated using the interatomic force constant $(k)$ and the mean interatomic distance $(r_0)$ by the formula: $Y = \frac{k}{r_0}$.
Given:
Interatomic force constant $k = 7 \ N/m$.
Mean interatomic distance $r_0 = 3 \times 10^{-10} \ m$.
Substituting the values:
$Y = \frac{7}{3 \times 10^{-10}} \ N/m^2$.
$Y = 2.333... \times 10^{10} \ N/m^2$.
Rounding to two decimal places,we get $Y = 2.33 \times 10^{10} \ N/m^2$.

(D) Given: Young's modulus $Y = 2.0 \times 10^{11} \ N/m^2$,Area $A = 0.1 \ cm^2 = 0.1 \times 10^{-4} \ m^2 = 10^{-5} \ m^2$.
To double the length of the wire,the change in length $\Delta L$ must be equal to the original length $L$,so $\Delta L = L$.
Strain is defined as $\frac{\Delta L}{L} = \frac{L}{L} = 1$.
Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$.
Substituting the values: $2.0 \times 10^{11} = \frac{F}{10^{-5} \times 1}$.
$F = 2.0 \times 10^{11} \times 10^{-5} \ N = 2 \times 10^6 \ N$.

(C) Young's modulus $(Y)$ is a characteristic property of the material of the wire.
It depends only on the nature of the material and the temperature.
It does not depend on the dimensions of the wire,such as its length $(L)$ or radius $(r)$.
Therefore,if the length and radius are doubled,the value of $Y$ remains the same.

(B) From Hooke's law,Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}}$.
The strain produced in a perfectly rigid body is zero,i.e.,$\text{Strain} = 0$.
Therefore,$Y = \frac{\text{Stress}}{0} \implies Y = \infty$.
Thus,the Young's modulus of a perfectly rigid body material is infinite.

(D) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since the volume $V = A \times L$ is constant,we can write $A = \frac{V}{L}$.
Substituting this into the elongation formula: $\Delta L = \frac{FL}{(V/L)Y} = \frac{FL^2}{VY}$.
Since $F$,$V$,and $Y$ are constant for both wires,we have $\Delta L \propto L^2$.
Given $\Delta L_1 = 2\, mm$ for $L_1 = 2\, m$,and we need to find $\Delta L_2$ for $L_2 = 8\, m$.
Using the ratio: $\frac{\Delta L_2}{\Delta L_1} = \left( \frac{L_2}{L_1} \right)^2 = \left( \frac{8}{2} \right)^2 = 4^2 = 16$.
Therefore,$\Delta L_2 = 16 \times \Delta L_1 = 16 \times 2\, mm = 32\, mm$.
Converting to centimeters: $32\, mm = 3.2\, cm$.

(A) The formula for the extension $(l)$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since the material $(Y)$,original length $(L)$,and applied force $(F)$ are constant,we have $l \propto \frac{1}{A}$.
Given $A_1 = 4 \; mm^2$,$l_1 = 0.1 \; mm$,and $A_2 = 8 \; mm^2$.
Using the ratio $\frac{l_2}{l_1} = \frac{A_1}{A_2}$,we get $\frac{l_2}{0.1} = \frac{4}{8} = \frac{1}{2}$.
Therefore,$l_2 = \frac{0.1}{2} = 0.05 \; mm$.

(A) The elongation $\Delta L$ of a rod of length $L$,density $d$,and Young's modulus $Y$ due to its own weight is given by the formula: $\Delta L = \frac{L^2 dg}{2Y}$.
Given values are: $L = 10\, m$,$d = 1500\, kg/m^3$,$Y = 5 \times 10^8\, N/m^2$,and $g = 10\, m/s^2$.
Substituting these values into the formula:
$\Delta L = \frac{(10)^2 \times 1500 \times 10}{2 \times 5 \times 10^8}$
$\Delta L = \frac{100 \times 15000}{10 \times 10^8}$
$\Delta L = \frac{1500000}{10^9} = 15 \times 10^5 \times 10^{-9} = 15 \times 10^{-4}\, m$.

(B) The extension $l$ in a rod is given by the formula $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,$r$ is the radius,and $Y$ is Young's modulus.
Since the rods are identical ($Y$ is constant) and the force $F$ is the same,the extension $l$ is proportional to $\frac{L}{r^2}$ or $\frac{L}{D^2}$ (where $D$ is the diameter).
Calculating the ratio $\frac{L}{D^2}$ for each option:
$(a)$ $\frac{10}{1^2} = 10$
$(b)$ $\frac{100}{2^2} = \frac{100}{4} = 25$
$(c)$ $\frac{200}{3^2} = \frac{200}{9} \approx 22.22$
$(d)$ $\frac{300}{4^2} = \frac{300}{16} = 18.75$
Comparing these values,the ratio is maximum for option $(b)$.

(A) The energy per unit volume $(E)$ stored in a stretched wire is given by the formula:
$E = \frac{1}{2} \times Y \times (\text{strain})^2$
Rearranging the formula to solve for the strain:
$(\text{strain})^2 = \frac{2E}{Y}$
Taking the square root of both sides:
$\text{strain} = \sqrt{\frac{2E}{Y}}$
Thus,the correct option is $A$.

(A) The elastic potential energy per unit volume $(u)$ is given by the formula: $u = \frac{\text{stress}^2}{2Y}$,where $Y$ is the Young's modulus.
Given that the stress applied on both wires is the same,we have $u \propto \frac{1}{Y}$.
Therefore,the ratio of elastic energy per unit volume for the two wires is:
$\frac{u_1}{u_2} = \frac{Y_2}{Y_1}$.
Given the ratio of Young's modulus is $\frac{Y_1}{Y_2} = \frac{2}{3}$,we substitute this into the equation:
$\frac{u_1}{u_2} = \frac{3}{2}$.
Thus,the ratio of elastic energy per unit volume is $3:2$.

(B) Consider a small element of length $dx$ at a distance $x$ from the free end of the rubber cord.
The weight of the portion below this element is $W = (A \cdot x \cdot D) \cdot g$.
The stress at this section is $\sigma = \frac{W}{A} = x D g$.
By Hooke's Law,the strain is $\epsilon = \frac{\sigma}{E} = \frac{x D g}{E}$.
The extension of the small element $dx$ is $dl = \epsilon dx = \frac{x D g}{E} dx$.
To find the total extension $l$,we integrate from $x = 0$ to $x = L$:
$l = \int_{0}^{L} \frac{D g}{E} x dx = \frac{D g}{E} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{L^2 D g}{2 E}$.

(C) The thermal stress $\sigma$ developed in a rod fixed between two rigid walls when heated by a temperature change $\Delta \theta$ is given by the formula:
$\sigma = Y \alpha \Delta \theta$
where $Y$ is the Young's modulus and $\alpha$ is the coefficient of linear expansion.
Given that the thermal stresses are equal $(\sigma_1 = \sigma_2)$ and the increase in temperature $\Delta \theta$ is the same for both rods,we have:
$Y_1 \alpha_1 \Delta \theta = Y_2 \alpha_2 \Delta \theta$
$Y_1 \alpha_1 = Y_2 \alpha_2$
Rearranging the terms to find the ratio $Y_1 : Y_2$:
$\frac{Y_1}{Y_2} = \frac{\alpha_2}{\alpha_1}$
Given $\alpha_1 : \alpha_2 = 2 : 3$,we have $\frac{\alpha_1}{\alpha_2} = \frac{2}{3}$,which implies $\frac{\alpha_2}{\alpha_1} = \frac{3}{2}$.
Therefore,$\frac{Y_1}{Y_2} = \frac{3}{2}$ or $Y_1 : Y_2 = 3 : 2$.

(C) The thermal stress developed in a rod prevented from expanding is given by the formula: $\sigma = Y \alpha \Delta \theta$,where $Y$ is Young's modulus,$\alpha$ is the coefficient of linear expansion,and $\Delta \theta$ is the change in temperature.
From this formula,it is clear that the stress depends on the material properties ($Y$ and $\alpha$) and the temperature change $(\Delta \theta)$.
It does not depend on the original length $(L)$ or the cross-sectional area $(A)$ of the rod.
Therefore,the stress is independent of the length of the rod.

(A) The ring is heated to fit over the wheel. When it cools down,it exerts a tension force $F$ on the wheel.
The thermal strain produced by the temperature change $\Delta T$ is given by $\epsilon = \alpha \Delta T$.
From the definition of Young's modulus,$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/S}{\Delta L/L}$.
Here,the strain $\frac{\Delta L}{L}$ is equal to $\alpha \Delta T$.
Therefore,the tension force $F$ in the ring is $F = Y S \alpha \Delta T$.
Since the ring is circular and exerts force on two semicircular parts,consider a small element of the ring subtending an angle $d\theta$ at the center. The radial force component is $dF_r = F d\theta$. Integrating this over a semicircle (from $0$ to $\pi$),the total force exerted by one half of the ring on the other half is $2F$.
However,the force exerted by one part of the wheel on the other part is balanced by the tension in the ring. The force $F$ calculated above is the tension in the ring. The force exerted by one semicircular part on the other is $2F_{radial} = 2F / \pi$ (for a full ring),but for this specific setup,the force exerted by one part of the wheel on the other is simply the tension $F$ acting at the two contact points. Thus,the force is $F = SY \alpha \Delta T$.

(B) Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta L} = \frac{MgL}{\pi (d/2)^2 \Delta L} = \frac{4MgL}{\pi d^2 \Delta L}$.
Given: $M = 10 \, kg$,$L = 2 \, m$,$g = 9.8 \, m/s^2$,$d = 0.4 \times 10^{-3} \, m$,$\Delta d = 0.01 \times 10^{-3} \, m$,$\Delta L = 0.88 \times 10^{-3} \, m$,$\delta(\Delta L) = 0.05 \times 10^{-3} \, m$.
Calculating $Y$: $Y = \frac{4 \times 10 \times 9.8 \times 2}{3.14 \times (0.4 \times 10^{-3})^2 \times 0.88 \times 10^{-3}} = \frac{784}{3.14 \times 0.16 \times 10^{-6} \times 0.88 \times 10^{-3}} \approx 1.78 \times 10^{11} \approx 2.0 \times 10^{11} \, N/m^2$.
Relative error: $\frac{\Delta Y}{Y} = 2 \frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L} = 2 \left( \frac{0.01}{0.4} \right) + \frac{0.05}{0.88} = 0.05 + 0.0568 \approx 0.1068$.
Absolute error: $\Delta Y = 0.1068 \times 2.0 \times 10^{11} \approx 0.21 \times 10^{11} \approx 0.2 \times 10^{11} \, N/m^2$.
Thus,$Y = (2.0 \pm 0.2) \times 10^{11} \, N/m^2$.

(A) Given: Diameter $d = 4 \, mm$,so radius $r = 2 \times 10^{-3} \, m$.
Young's modulus $Y = 9 \times 10^{10} \, N/m^2$.
Strain $\frac{l}{L} = 0.1\% = \frac{0.1}{100} = 0.001$.
The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot l}$,where $A = \pi r^2$.
Rearranging for force: $F = Y \cdot A \cdot \frac{l}{L}$.
Substituting the values: $F = (9 \times 10^{10}) \times (\pi \times (2 \times 10^{-3})^2) \times 0.001$.
$F = 9 \times 10^{10} \times \pi \times 4 \times 10^{-6} \times 10^{-3}$.
$F = 36 \times 10^1 \times \pi = 360 \pi \, N$.

(D) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$.
Since volume $V = A \times L$,we can write $A = \frac{V}{L}$.
Substituting this into the formula,we get $l = \frac{FL}{(V/L)Y} = \frac{FL^2}{VY}$.
Given that $V$,$Y$,and $F$ are constant,we have $l \propto L^2$.
Therefore,$\frac{l_2}{l_1} = \left( \frac{L_2}{L_1} \right)^2$.
Given $L_1 = 2 \, m$,$L_2 = 8 \, m$,and $l_1 = 2 \, mm = 0.2 \, cm$.
$\frac{l_2}{0.2} = \left( \frac{8}{2} \right)^2 = (4)^2 = 16$.
$l_2 = 16 \times 0.2 \, cm = 3.2 \, cm$.

(A) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{A Y} = \frac{FL}{\pi r^2 Y}$,where $Y$ is Young's modulus.
For the first wire: $l_1 = \frac{F_1 L_1}{\pi r_1^2 Y} = \frac{FL}{\pi r^2 Y} = l$.
For the second wire: $l_2 = \frac{F_2 L_2}{\pi r_2^2 Y} = \frac{(2F)(2L)}{\pi (2r)^2 Y} = \frac{4FL}{\pi (4r^2) Y} = \frac{FL}{\pi r^2 Y}$.
Comparing the two,we get $l_2 = l_1 = l$.

(D) The Young's modulus $Y$ is given by the formula $Y = \frac{F L}{A \Delta l}$,where $A = \pi r^2$ is the cross-sectional area.
Rearranging for force $F$,we get $F = \frac{Y A \Delta l}{L} = \frac{Y \pi r^2 \Delta l}{L}$.
Given that the material is the same,$Y_A = Y_B$. Also,the extension $\Delta l$ is the same for both wires,so $\Delta l_A = \Delta l_B$.
The ratio of forces is $\frac{F_A}{F_B} = \frac{Y_A \pi r_A^2 \Delta l_A / L_A}{Y_B \pi r_B^2 \Delta l_B / L_B} = \left( \frac{r_A}{r_B} \right)^2 \left( \frac{L_B}{L_A} \right)$.
Given $\frac{L_A}{L_B} = \frac{1}{2}$ and $\frac{r_A}{r_B} = \frac{2}{1}$.
Substituting these values: $\frac{F_A}{F_B} = (2)^2 \times (2) = 4 \times 2 = 8$.
Thus,the ratio $\frac{F_A}{F_B} = 8:1$.

(B) According to Hooke's Law,the length $L$ of a wire under tension $F$ is given by $L_{total} = L_0 + \frac{F}{k}$,where $L_0$ is the natural length and $k$ is the force constant.
For $F = 4 \, N$,$L_0 + \frac{4}{k} = a$ --- $(i)$
For $F = 5 \, N$,$L_0 + \frac{5}{k} = b$ --- $(ii)$
Subtracting $(i)$ from $(ii)$:
$\frac{5}{k} - \frac{4}{k} = b - a \implies \frac{1}{k} = b - a$.
Substituting $\frac{1}{k}$ into $(i)$:
$L_0 + 4(b - a) = a \implies L_0 = a - 4b + 4a = 5a - 4b$.
Now,for $F = 9 \, N$:
$L_{total} = L_0 + \frac{9}{k} = (5a - 4b) + 9(b - a)$.
$L_{total} = 5a - 4b + 9b - 9a = 5b - 4a$.

(D) The change in length $\Delta L$ is given by the formula: $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi (d/2)^2 Y} = \frac{4FL}{\pi d^2 Y}$.
Since the force $F$,Young's modulus $Y$,and $\pi$ are constant for all wires,we have $\Delta L \propto \frac{L}{d^2}$.
For option $A$: $\frac{100}{1^2} = 100$.
For option $B$: $\frac{200}{2^2} = \frac{200}{4} = 50$.
For option $C$: $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$.
For option $D$: $\frac{50}{0.5^2} = \frac{50}{0.25} = 200$.
Comparing the values,the ratio is maximum for option $D$ $(200 > 100 > 50 > 33.33)$.
Therefore,the increase in length is maximum for the wire in option $D$.

(A) Young's modulus $(Y)$ is a material property that depends only on the nature of the material,not on the dimensions of the object (length,radius,etc.).
Since both wires are made of the same material,their Young's modulus will be identical.
Therefore,the Young's modulus of the second wire remains $Y$.

(A) When a wire is fixed at both ends (or prevented from expanding),the thermal stress developed due to a change in temperature $\Delta \theta$ is given by the formula: $\sigma = Y \alpha \Delta \theta$.
Here,$Y = 1.2 \times 10^{11} \ N/m^2$,$\alpha = 1.1 \times 10^{-5} \ ^oC^{-1}$,and $\Delta \theta = 10^oC$.
Substituting the values:
$\sigma = (1.2 \times 10^{11}) \times (1.1 \times 10^{-5}) \times 10$
$\sigma = 1.32 \times 10^{11} \times 10^{-4}$
$\sigma = 1.32 \times 10^7 \ N/m^2$.
Note: The applied force of $200 \ N$ is irrelevant to the calculation of thermal stress in this context as the question asks for the stress due to temperature change.

(A) Given:
Cross-sectional area $A = 1 \, cm^2 = 1 \times 10^{-4} \, m^2$.
Young's modulus $Y = 2 \times 10^{11} \, N/m^2$.
Final length $L_2 = 1.1 \, L_1$.
Strain is defined as $\frac{\Delta L}{L_1} = \frac{L_2 - L_1}{L_1} = \frac{1.1 \, L_1 - L_1}{L_1} = 0.1$.
Using the formula for Young's modulus: $Y = \frac{F \cdot L_1}{A \cdot \Delta L}$,we can rearrange to find the force $F$:
$F = Y \cdot A \cdot \left( \frac{\Delta L}{L_1} \right)$.
Substituting the values:
$F = (2 \times 10^{11}) \times (1 \times 10^{-4}) \times (0.1)$.
$F = 2 \times 10^{11} \times 10^{-4} \times 10^{-1}$.
$F = 2 \times 10^{6} \, N$.

(B) In the first case,the wire is suspended from a rigid support with a weight $W$ at its free end. The tension in the wire is $T_1 = W$. The extension is given by $\Delta l_1 = \frac{T_1 L}{AY} = \frac{WL}{AY} = 1.0 \, mm$.
In the second case,the wire is passed over a massless,frictionless pulley with a weight $W$ attached to each end. The tension in the wire throughout its length is $T_2 = W$.
The extension in this case is $\Delta l_2 = \frac{T_2 L}{AY} = \frac{WL}{AY}$.
Since $T_1 = T_2 = W$,the extension remains the same.
Therefore,$\Delta l_2 = 1.0 \, mm$.

(D) The formula for the extension in a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Assuming the original lengths $L$ are equal for all wires,the extension $\Delta L$ is proportional to $\frac{1}{AY}$.
Given the ratios: $Y_1 : Y_2 : Y_3 = 2 : 2 : 1$ and $A_1 : A_2 : A_3 = 1 : 2 : 3$.
The ratio of extensions is $\Delta L_1 : \Delta L_2 : \Delta L_3 = \frac{1}{A_1 Y_1} : \frac{1}{A_2 Y_2} : \frac{1}{A_3 Y_3}$.
Substituting the values: $\Delta L_1 : \Delta L_2 : \Delta L_3 = \frac{1}{1 \times 2} : \frac{1}{2 \times 2} : \frac{1}{3 \times 1}$.
This simplifies to $\frac{1}{2} : \frac{1}{4} : \frac{1}{3}$.
To clear the denominators,multiply by the least common multiple $(12)$: $\frac{12}{2} : \frac{12}{4} : \frac{12}{3} = 6 : 3 : 4$.

(C) The displacement of point $B$ is equal to the elongation of the wire segment $AB$.
The formula for elongation $\Delta L$ is given by $\Delta L = \frac{MgL}{AY}$,where $M$ is the mass,$g$ is the acceleration due to gravity,$L$ is the length of the wire,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
For segment $AB$:
$M = 10 \, kg$
$g = 10 \, m/s^2$
$L = 0.1 \, m$
$A = 10^{-4} \, m^2$
$Y = 2.5 \times 10^{10} \, N/m^2$
Substituting these values:
$\Delta L_{AB} = \frac{10 \times 10 \times 0.1}{10^{-4} \times 2.5 \times 10^{10}}$
$\Delta L_{AB} = \frac{10}{2.5 \times 10^6} = 4 \times 10^{-6} \, m$
Therefore,the displacement of point $B$ is $4 \times 10^{-6} \, m$.

(A) The displacement of point $D$ is the sum of the elongations of segments $AB$,$BC$,and $CD$. The force $F = Mg = 10 \times 10 = 100 \ N$ acts on all segments.
Elongation $\Delta L = \frac{FL}{AY}$.
For segment $AB$: $\Delta L_{AB} = \frac{100 \times 0.1}{10^{-4} \times 2.5 \times 10^{10}} = \frac{10}{2.5 \times 10^6} = 4 \times 10^{-6} \ m$.
For segment $BC$: $\Delta L_{BC} = \frac{100 \times 0.2}{10^{-4} \times 4 \times 10^{10}} = \frac{20}{4 \times 10^6} = 5 \times 10^{-6} \ m$.
For segment $CD$: $\Delta L_{CD} = \frac{100 \times 0.15}{10^{-4} \times 1 \times 10^{10}} = \frac{15}{1 \times 10^6} = 15 \times 10^{-6} \ m$.
Total displacement of $D = \Delta L_{AB} + \Delta L_{BC} + \Delta L_{CD} = (4 + 5 + 15) \times 10^{-6} \ m = 24 \times 10^{-6} \ m$.

(B) For a system of two masses $m_1$ and $m_2$ connected by a string over a pulley,the tension $T$ in the string is given by:
$T = \frac{{2{m_1}{m_2}}}{{{m_1} + {m_2}}}g$
The stress in the string is defined as the force per unit area:
$\text{Stress} = \frac{T}{S} = \frac{{2{m_1}{m_2}g}}{{S({m_1} + {m_2})}}$
According to Hooke's Law,Young's modulus $Y$ is the ratio of stress to strain:
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\text{Stress}}{l/L}$
Rearranging for elongation $l$:
$l = \frac{\text{Stress} \cdot L}{Y} = \frac{{2{m_1}{m_2}g}}{{S({m_1} + {m_2})}} \cdot \frac{L}{Y}$
Therefore,the elongation $l$ is:
$l = \frac{{2{m_1}{m_2}gL}}{{YS({m_1} + {m_2})}}$

(C) For the equilibrium of the mass $m$,the vertical component of the tension $T$ in the wire balances the weight:
$2T \sin \theta = mg$
Since $x << L$,$\sin \theta \approx \tan \theta = \frac{x}{L}$.
Thus,$T = \frac{mg}{2 \sin \theta} = \frac{mgL}{2x}$.
The extension in the wire $\Delta L$ is given by the difference between the new length of the wire and its original length:
$\Delta L = 2 \sqrt{L^2 + x^2} - 2L = 2L \left[ (1 + \frac{x^2}{L^2})^{1/2} - 1 \right]$
Using the binomial expansion $(1+z)^n \approx 1+nz$ for $z << 1$:
$\Delta L \approx 2L \left[ 1 + \frac{1}{2} \frac{x^2}{L^2} - 1 \right] = \frac{x^2}{L}$.
From Young's modulus $Y = \frac{T/A}{\Delta L / L_{original}}$,where $L_{original} = L$ for each half of the wire:
$T = \frac{YA \Delta L}{L} = \frac{YA (x^2/L)}{L} = \frac{YAx^2}{L^2}$.
Equating the two expressions for $T$:
$\frac{mgL}{2x} = \frac{YAx^2}{L^2}$
$mg = \frac{2YAx^3}{L^3}$
$m = \frac{2YAx^3}{gL^3}$.
Note: Given the options provided,the closest form is $m = \frac{YAx^3}{gL^3}$ (Option $C$),assuming the total length $2L$ is treated as the original length for the modulus calculation.

(B) The force constant $k$ of a wire is given by $k = \frac{YA}{L}$.
Since the two wires are in parallel,the total force constant $k_{eq}$ is the sum of the individual force constants: $k_{eq} = k_1 + k_2$.
Substituting the expressions for $k_1$ and $k_2$: $\frac{Y_{eq} (2A)}{L} = \frac{Y_1 A}{L} + \frac{Y_2 A}{L}$.
Here,the equivalent system has a total cross-sectional area of $2A$ and the same length $L$.
Simplifying the equation: $Y_{eq} (2A) = (Y_1 + Y_2) A$.
Therefore,$Y_{eq} = \frac{Y_1 + Y_2}{2}$.

(D) The thermal strain produced in the wire due to a temperature change $T$ is given by $\frac{\Delta L}{L} = \alpha T$,where $\alpha$ is the linear expansion coefficient.
Since the volume expansion coefficient $\gamma = 3\alpha$,we have $\alpha = \frac{\gamma}{3}$.
Thus,the strain is $\frac{\Delta L}{L} = \frac{\gamma T}{3}$.
The energy density $u$ (energy per unit volume) stored in a stretched wire is given by $u = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.
Using Hooke's Law,$\text{Stress} = Y \times \text{Strain}$,so $u = \frac{1}{2} Y (\text{Strain})^2$.
Substituting the value of strain: $u = \frac{1}{2} Y \left( \frac{\gamma T}{3} \right)^2$.
$u = \frac{1}{2} Y \left( \frac{\gamma^2 T^2}{9} \right) = \frac{1}{18} \gamma^2 T^2 Y$.

(B) The Young's modulus $Y$ is defined as $Y = \frac{FL}{A\Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.
Rearranging for the change in length,we get $\Delta L = \frac{4FL}{\pi D^2 Y}$.
Let the subscripts $S$ and $C$ refer to steel and copper wires respectively.
The force on the steel wire is $F_S = (5m + 2m)g = 7mg$.
The force on the copper wire is $F_C = 5mg$.
Given ratios are $\frac{D_S}{D_C} = p$,$\frac{L_S}{L_C} = q$,and $\frac{Y_S}{Y_C} = s$.
The ratio of increase in their lengths is $\frac{\Delta L_S}{\Delta L_C} = \left( \frac{F_S}{F_C} \right) \left( \frac{L_S}{L_C} \right) \left( \frac{D_C}{D_S} \right)^2 \left( \frac{Y_C}{Y_S} \right)$.
Substituting the given values: $\frac{\Delta L_S}{\Delta L_C} = \left( \frac{7mg}{5mg} \right) (q) \left( \frac{1}{p} \right)^2 \left( \frac{1}{s} \right) = \frac{7q}{5p^2s}$.

(A) Young's modulus is given by $Y = \frac{FL}{A \Delta L} = \frac{4FL}{\pi D^2 \Delta L}$.
Rearranging for extension,we get $\Delta L = \frac{4FL}{\pi D^2 Y}$.
Since all wires are made of the same material and subjected to the same tension $F$,$Y$ and $F$ are constant.
Therefore,$\Delta L \propto \frac{L}{D^2}$.
Calculating the ratio $\frac{L}{D^2}$ for each case:
$(a)$ $\frac{50}{(0.5)^2} = \frac{50}{0.25} = 200 \; cm^{-1}$
$(b)$ $\frac{100}{(1)^2} = 100 \; cm^{-1}$
$(c)$ $\frac{200}{(2)^2} = \frac{200}{4} = 50 \; cm^{-1}$
$(d)$ $\frac{300}{(3)^2} = \frac{300}{9} \approx 33.3 \; cm^{-1}$
Comparing the values,the ratio is largest for option $(a)$.

(B) Given that the volume $V$ of the copper wire is constant,we have $V = A \cdot l$,where $A$ is the cross-sectional area and $l$ is the length of the wire.
From this,we can express the area as $A = \frac{V}{l}$.
According to Hooke's Law,Young's modulus $Y$ is defined as $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{A \cdot \Delta l}$.
Rearranging this formula to solve for the extension $\Delta l$,we get $\Delta l = \frac{F \cdot l}{Y \cdot A}$.
Substituting $A = \frac{V}{l}$ into the equation,we get $\Delta l = \frac{F \cdot l}{Y \cdot (V/l)} = \frac{F \cdot l^2}{Y \cdot V}$.
Since $F$,$Y$,and $V$ are constants,we have $\Delta l \propto l^2$.
Therefore,the graph of $\Delta l$ versus $l^2$ will be a straight line passing through the origin.

(A) Let $L$ and $A$ be the length and area of cross-section of each wire respectively.
In order to have the lower ends of the wires at the same level,the elongation produced in both wires must be equal.
Let $W_s$ and $W_b$ be the weights added to the steel and brass wires respectively.
By the definition of Young's modulus $Y = \frac{W/A}{\Delta L/L}$,the elongation produced in the steel wire is $\Delta L_s = \frac{W_s L}{Y_s A}$.
The elongation produced in the brass wire is $\Delta L_b = \frac{W_b L}{Y_b A}$.
Since $\Delta L_s = \Delta L_b$,we have $\frac{W_s L}{Y_s A} = \frac{W_b L}{Y_b A}$.
This simplifies to $\frac{W_s}{W_b} = \frac{Y_s}{Y_b}$.
Given that the Young's modulus of steel is twice that of brass,$Y_s = 2 Y_b$,so $\frac{Y_s}{Y_b} = 2$.
Therefore,$\frac{W_s}{W_b} = 2$,which means the ratio is $2:1$.

(A) Young's modulus is given by $Y = \frac{Fl}{A\Delta l}$.
Since the volume $V = A \times L$ is the same for both wires,and the cross-sectional areas are $A_1 = A$ and $A_2 = 3A$,their lengths must be $L_1 = 3l$ and $L_2 = l$ respectively.
For the first wire:
$\Delta l = \frac{F \cdot (3l)}{A \cdot Y} = \frac{3Fl}{AY} \quad ...(i)$
For the second wire,let the required force be $F'$. The extension is the same $\Delta l$:
$\Delta l = \frac{F' \cdot l}{(3A) \cdot Y} = \frac{F'l}{3AY} \quad ...(ii)$
Equating $(i)$ and $(ii)$:
$\frac{3Fl}{AY} = \frac{F'l}{3AY}$
$3F = \frac{F'}{3}$
$F' = 9F$