Young’s Modulus Questions in English

(A) The formula for Young's modulus $(Y)$ is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}}$.
Since strain is a dimensionless quantity (ratio of change in length to original length),the units of Young's modulus are the same as the units of stress.
Stress is defined as force per unit area: $\text{Stress} = \frac{\text{Force}}{\text{Area}}$.
Pressure is also defined as force per unit area: $\text{Pressure} = \frac{\text{Force}}{\text{Area}}$.
Therefore,Young's modulus has the same units as pressure,which is $N/m^2$ or Pascal $(Pa)$.

(A) Young's modulus $(Y)$ is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}}$.
Since strain is a dimensionless quantity,the unit of Young's modulus is the same as the unit of stress.
Stress is defined as force per unit area,so its $SI$ unit is $N m^{-2}$ (or Pascal,$Pa$) and its $CGS$ unit is $Dyne cm^{-2}$.
$N m^{-1}$ is the unit of force constant or surface tension,not stress or Young's modulus.
Therefore,$N m^{-1}$ is not a unit of Young's modulus.

(B) Young's modulus $(Y)$ is defined as the ratio of stress to strain:
$Y = \frac{\text{Stress}}{\text{Strain}}$
Stress is defined as force per unit area,so its unit is $N/m^2$ or $N m^{-2}$.
Strain is a dimensionless quantity because it is the ratio of change in dimension to the original dimension.
Therefore,the unit of Young's modulus is the same as the unit of stress,which is $N m^{-2}$.

(A) Young's modulus $Y$ is defined as the ratio of stress to strain.
$Y = \frac{\text{Stress}}{\text{Strain}}$
Since strain is a dimensionless quantity,the dimensions of Young's modulus are the same as those of stress.
Stress = $\frac{\text{Force}}{\text{Area}} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$
Therefore,the dimensional formula for Young's modulus is $[M L^{-1} T^{-2}]$.

(D) The spring constant $k$ of a spring is inversely proportional to its natural length $l$,given by the relation $k \propto \frac{1}{l}$.
This is because for a given material and cross-sectional area,the stiffness of a spring decreases as its length increases.
Mathematically,if $Y$ is Young's modulus and $A$ is the cross-sectional area,$k = \frac{YA}{l}$.
Since $Y$ and $A$ are constants for a given spring,we have $k \propto \frac{1}{l}$.
The graph of $k$ versus $l$ represents a rectangular hyperbola,which is correctly depicted in graph $D$.

(C) The formula for Young's modulus $(Y)$ is given by $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} = \frac{FL}{Al}$.
Rearranging this formula to solve for the increase in length $(l)$,we get:
$l = \frac{FL}{YA}$.
From this expression,it is clear that the increase in length $(l)$ is directly proportional to the force $(F)$ and the original length $(L)$,and inversely proportional to the Young's modulus $(Y)$ and the area of cross-section $(A)$.
Therefore,$l \propto \frac{1}{A}$.
Thus,the correct statement is that the increase in length is inversely proportional to $A$.

(D) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$,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 $\Delta L$,we get $\Delta L = \frac{F \cdot L}{A \cdot Y}$.
Since the material is the same,$Y$ is constant. For the same tension $F$,we have $\Delta L \propto \frac{L}{A}$.
Since $A = \pi \cdot r^2 = \pi \cdot (d/2)^2 = \frac{\pi \cdot d^2}{4}$,we have $\Delta L \propto \frac{L}{d^2}$.
Calculating the ratio $\frac{L}{d^2}$ for each option:
$(a)$ $\frac{100}{1^2} = 100$
$(b)$ $\frac{200}{2^2} = \frac{200}{4} = 50$
$(c)$ $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$
$(d)$ $\frac{50}{0.5^2} = \frac{50}{0.25} = 200$
Since the ratio is maximum for option $(d)$,the increase in length is maximum for wire $(d)$.

(B) The Young's modulus $(Y)$ is an intrinsic property of the material of the wire.
It depends only on the nature of the material and the temperature,not on the dimensions of the wire such as its length $(L)$ or radius $(r)$.
Therefore,even if the length and radius are changed,the Young's modulus remains constant.
Thus,the new Young's modulus will be $Y$.

(C) The depression $\delta$ at the centre of a beam of length $L$,breadth $b$,and depth $d$,supported at its two ends and loaded with a weight $W$ at the centre,is given by the formula:
$\delta = \frac{WL^3}{4Ybd^3}$
Where $Y$ is the Young's modulus of the material of the beam.
From this relation,it is clear that the depression $\delta$ is inversely proportional to the Young's modulus $Y$ of the material.
Therefore,$\delta \propto \frac{1}{Y}$.

(C) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$.
Given that $F$,$L$,and $Y$ are constant,the elongation $l$ is inversely proportional to the square of the diameter: $l \propto \frac{1}{d^2}$.
Let $l_1 = 1 \ cm$ and $d_1 = d$. For the second wire,$d_2 = d/2$.
Using the ratio: $\frac{l_2}{l_1} = \left( \frac{d_1}{d_2} \right)^2 = \left( \frac{d}{d/2} \right)^2 = (2)^2 = 4$.
Therefore,$l_2 = 4 \times l_1 = 4 \times 1 \ cm = 4 \ cm$.

(B) The Young's modulus $Y$ is given by the formula $Y = \frac{F L}{A \Delta L}$,where $A = \pi r^2$.
Rearranging for the change in length $\Delta L$,we get $\Delta L = \frac{F L}{\pi r^2 Y}$.
Since the force $F$,original length $L$,and Young's modulus $Y$ remain constant,the relationship is $\Delta L \propto \frac{1}{r^2}$.
Given the initial radius is $r_1$ and the new radius is $r_2 = 2r_1$,the new change in length $\Delta L_2$ is related to the initial change $\Delta L_1 = 12 \ mm$ by:
$\frac{\Delta L_2}{\Delta L_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{2r_1} \right)^2 = \frac{1}{4}$.
Therefore,$\Delta L_2 = \frac{\Delta L_1}{4} = \frac{12 \ mm}{4} = 3 \ mm$.

(C) Given:
Length of wire $L = 1.1 \, m$
Area of cross-section $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$
Mass $m = 1 \, kg$
Young's modulus $Y = 1.1 \times 10^{11} \, N/m^2$
Acceleration due to gravity $g = 10 \, m/s^2$
The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F = mg$.
Rearranging for the change in length $\Delta L$:
$\Delta L = \frac{mgL}{AY}$
Substituting the values:
$\Delta L = \frac{1 \times 10 \times 1.1}{(1 \times 10^{-6}) \times (1.1 \times 10^{11})}$
$\Delta L = \frac{11}{1.1 \times 10^5} = \frac{10}{10^5} = 10^{-4} \, m$
Converting to $mm$:
$\Delta L = 10^{-4} \times 10^3 \, mm = 0.1 \, mm$.

(D) Given:
Length of wire $L = 2\, m$
Change in length $\Delta L = 0.5\, mm = 0.5 \times 10^{-3}\, m$
Area of cross-section $A = 2\, mm^2 = 2 \times 10^{-6}\, m^2$
Young's modulus $Y = 2.2 \times 10^{11}\, N/m^2$
Formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$.
Rearranging for force $F$:
$F = \frac{Y \cdot A \cdot \Delta L}{L}$
Substituting the values:
$F = \frac{(2.2 \times 10^{11}) \times (2 \times 10^{-6}) \times (0.5 \times 10^{-3})}{2}$
$F = \frac{2.2 \times 10^{11} \times 1 \times 10^{-9}}{2}$
$F = \frac{2.2 \times 10^2}{2} = 1.1 \times 10^2\, N$.

(A) The interatomic force constant $K$ is related to Young's modulus $Y$ and the interatomic spacing $r_0$ by the formula $K = Y \times r_0$.
Given:
Young's modulus $Y = 2 \times 10^{11} \ N/m^2$.
Interatomic spacing $r_0 = 3 \times 10^{-10} \ m$.
Substituting these values into the formula:
$K = (2 \times 10^{11} \ N/m^2) \times (3 \times 10^{-10} \ m)$
$K = 6 \times 10^{1} \ N/m$
$K = 60 \ N/m$.
Therefore,the interatomic force constant is $60 \ N/m$.

(B) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.
Given that the length is to be doubled,the change in length $\Delta L = L$,so the strain $\frac{\Delta L}{L} = 1$.
Substituting the values,we get $Y = \frac{F/A}{1}$,which implies $F = Y \times A$.
Given $Y = 2 \times 10^{12} \text{ dyn/cm}^2$ and unit cross-sectional area $A = 1 \text{ cm}^2$.
Therefore,$F = (2 \times 10^{12} \text{ dyn/cm}^2) \times (1 \text{ cm}^2) = 2 \times 10^{12} \text{ dynes}$.

(A) Given: Diameter $d = 4 \ mm = 4 \times 10^{-3} \ m$.
Radius $r = 2 \times 10^{-3} \ m$.
Area $A = \pi r^2 = \pi (2 \times 10^{-3})^2 = 4\pi \times 10^{-6} \ m^2$.
Young's modulus $Y = 9 \times 10^{10} \ N/m^2$.
Strain $\frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 10^{-3}$.
Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$.
Rearranging for force: $F = Y \cdot A \cdot \left( \frac{\Delta L}{L} \right)$.
Substituting the values: $F = (9 \times 10^{10}) \times (4\pi \times 10^{-6}) \times (10^{-3})$.
$F = 36\pi \times 10^{10-6-3} = 36\pi \times 10^1 = 360\pi \ N$.

(C) The elongation $l$ of a wire under a load $F$ is given by the formula $l = \frac{FL}{AY},$ where $A$ is the cross-sectional area and $Y$ is Young's modulus.
Since $F,$ $A,$ and $Y$ are constant,the elongation $l$ is directly proportional to the original length $L$ of the wire $(l \propto L)$.
If the length $L$ is reduced to half (i.e.,$L' = \frac{L}{2}$),the new increase in length $l'$ will be $l' = \frac{Fl'}{AY} = \frac{F(L/2)}{AY} = \frac{1}{2} \left( \frac{FL}{AY} \right) = \frac{l}{2}.$

(B) The elongation $\Delta L$ of a string 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: $L = 8\, cm = 8 \times 10^{-2}\, m$,$d = 1.5\, kg/m^3$,$Y = 5 \times 10^8\, N/m^2$,and $g = 9.8\, m/s^2$.
Substituting these values into the formula:
$\Delta L = \frac{(8 \times 10^{-2})^2 \times 1.5 \times 9.8}{2 \times 5 \times 10^8}$
$\Delta L = \frac{64 \times 10^{-4} \times 14.7}{10^9}$
$\Delta L = \frac{940.8 \times 10^{-4}}{10^9} = 940.8 \times 10^{-13} = 9.408 \times 10^{-11}\, m$.
Rounding to the nearest provided option,the result is $9.6 \times 10^{-11}\, m$.

(D) The formula for Young's modulus is $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.
Given that the length is doubled,the change in length $\Delta L = L_{final} - L_{initial} = 2L - L = L$.
Therefore,the strain is $\frac{\Delta L}{L} = \frac{L}{L} = 1$.
Substituting the values into the formula: $Y = \frac{F/A}{1} \implies F = Y \times A$.
Given $Y = 10^{12} \, dyne/cm^2$ and $A = 0.5 \, cm^2$.
$F = 10^{12} \times 0.5 = 0.5 \times 10^{12} \, dyne$.

(D) The formula for the extension in a wire is given by $l = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since the wires are made of the same material,$Y$ is the same. Given that $F$ and $L$ are also identical,we have $l \propto \frac{1}{A}$.
Since $A = \pi r^2$,where $r$ is the radius,we get $l \propto \frac{1}{r^2}$.
Given that the diameter of wire $A$ is twice that of wire $B$,the radius $r_A = 2r_B$.
Therefore,$\frac{l_A}{l_B} = \left( \frac{r_B}{r_A} \right)^2 = \left( \frac{r_B}{2r_B} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
This implies $l_A = \frac{1}{4} l_B$,which means the increase in length for wire $A$ is one-fourth that of wire $B$.

(C) spring is considered better if it develops a large restoring force upon being deformed. This property depends on the elasticity of the material. Since the Young's modulus of elasticity of steel is greater than that of copper,steel is preferred for making springs.

(B) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.
Since the wires have the same length $(L)$,are stretched by the same weight $(F)$,and assuming they have the same cross-sectional area $(A)$,the extension $l$ is inversely proportional to the Young's modulus $(Y)$: $l \propto \frac{1}{Y}$.
Therefore,the ratio of the increase in length of steel $(l_S)$ to copper $(l_{Cu})$ is $\frac{l_S}{l_{Cu}} = \frac{Y_{Cu}}{Y_S}$.
Substituting the given values: $\frac{l_S}{l_{Cu}} = \frac{1.2 \times 10^{11}}{2 \times 10^{11}} = \frac{1.2}{2} = \frac{12}{20} = \frac{3}{5}$.

(B) Given:
Area of cross-section $A = 2 \, cm^2$
Force $F = 2 \times 10^5 \, dynes$
Initial length $= L$
Final length $= 2L$
Change in length $\Delta L = 2L - L = L$
Strain $= \frac{\Delta L}{L} = \frac{L}{L} = 1$
Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$
Substituting the values:
$Y = \frac{(2 \times 10^5) / 2}{1} = 10^5 \, dyne/cm^2$
Therefore,the correct option is $B$.

(B) The formula for the extension of a wire is given by $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since the weight $F$ and Young's modulus $Y$ remain constant,the extension $\Delta L$ is proportional to $\frac{L}{r^2}$.
Let the initial length be $L_1 = L$ and initial radius be $r_1 = r$. The initial extension is $\Delta L_1 = 1 \, mm$.
For the second wire,the length is $L_2 = 2L$ and the radius is $r_2 = 2r$.
Using the proportionality $\frac{\Delta L_2}{\Delta L_1} = \frac{L_2}{L_1} \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{\Delta L_2}{1} = \frac{2L}{L} \times \left( \frac{r}{2r} \right)^2 = 2 \times \left( \frac{1}{2} \right)^2 = 2 \times \frac{1}{4} = 0.5$.
Therefore,the increase in length is $0.5 \, mm$.

(C) When a rod is heated,it tends to expand. If it is prevented from expanding,a thermal stress is developed in the rod.
The thermal strain is given by $\frac{\Delta l}{l} = \alpha \Delta \theta$,where $\alpha$ is the coefficient of linear expansion and $\Delta \theta$ is the change in temperature.
From the definition of Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l}$.
Therefore,the force $F$ developed is $F = YA \left( \frac{\Delta l}{l} \right) = YA \alpha \Delta \theta$.
Since $Y$,$\alpha$,and $\Delta \theta$ are constant for the given rod,we have $F \propto A$.

(D) The formula for Young's modulus is $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{Strain}}$.
Rearranging for the cross-sectional area $A$,we get $A = \frac{F}{Y \times \text{Strain}}$.
Given: Load $F = 10^4 \, N$,Young's modulus $Y = 7 \times 10^9 \, N/m^2$,and breaking strain $= 0.2\% = 0.002$.
Substituting the values: $A = \frac{10^4}{7 \times 10^9 \times 0.002}$.
$A = \frac{10^4}{1.4 \times 10^7} = \frac{1}{1400} \approx 7.14 \times 10^{-4} \, m^2$.
Thus,the minimum cross-sectional area is $7.1 \times 10^{-4} \, m^2$.

(B) The longitudinal strain is given by $\text{strain} = \frac{\Delta L}{L} = \frac{\text{stress}}{Y} = \frac{F}{A \cdot Y}$,where $F$ is the force,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since both wires are made of copper,$Y$ is the same for both.
Given the same force $F$,the strain is inversely proportional to the area $A$: $\text{strain} \propto \frac{1}{A}$.
Since $A = \pi r^2$,we have $\text{strain} \propto \frac{1}{r^2}$.
Therefore,the ratio of strain is $\frac{\text{strain}_1}{\text{strain}_2} = \frac{r_2^2}{r_1^2} = \left( \frac{r_2}{r_1} \right)^2$.
Given $\frac{r_1}{r_2} = \frac{1}{4}$,we have $\frac{r_2}{r_1} = 4$.
Thus,the ratio of strain is $(4)^2 = 16$,which is $16:1$.

(A) Given: Force $F = mg = 200 \, kg \times 9.8 \, m/s^2 \approx 2000 \, N$ (taking $g = 10 \, m/s^2$ for standard calculation),
Original length $L = 600.5 \, cm - 0.5 \, cm = 600 \, cm = 6 \, m$,
Change in length $\Delta L = 0.5 \, cm = 0.5 \times 10^{-2} \, m$,
Area $A = 1 \, mm^2 = 10^{-6} \, m^2$.
Using the formula for Young's modulus $Y = \frac{F \cdot L}{A \cdot \Delta L}$:
$Y = \frac{2000 \times 6}{10^{-6} \times 0.5 \times 10^{-2}}$
$Y = \frac{12000}{0.5 \times 10^{-8}} = 24000 \times 10^8 = 2.4 \times 10^{12} \, N/m^2$.
Considering the provided options,the closest value is $2.35 \times 10^{12} \, N/m^2$.

(A) The formula for the extension $(l)$ of a wire is given by $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since the material is the same,$Y$ is constant. Given that the load $(F)$ and the length $(L)$ are also constant,we have $l \propto \frac{1}{r^2}$.
Let $d_1$ and $d_2$ be the diameters of the two wires. We are given $\frac{d_1}{d_2} = n : 1$,which implies $\frac{r_1}{r_2} = n$.
For the thin wire (wire $2$),the extension $l_2$ relative to the thick wire (wire $1$) is given by $\frac{l_2}{l_1} = \left( \frac{r_1}{r_2} \right)^2 = n^2$.
Therefore,$l_2 = n^2 l_1$.

(B) Given: Stress $\sigma = 1\,kg/mm^2 = 10^6\,kg/m^2$.
Converting to $N/m^2$ (taking $g = 10\,m/s^2$): $\sigma = 10^6 \times 10 = 10^7\,N/m^2$.
Young's Modulus $Y = 10^{11}\,N/m^2$.
Longitudinal strain $\frac{\Delta L}{L} = \frac{\sigma}{Y} = \frac{10^7}{10^{11}} = 10^{-4}$.
Percentage increase in length $= \frac{\Delta L}{L} \times 100 = 10^{-4} \times 100 = 0.01\%$.

(D) The Young's modulus $(Y)$ is an intrinsic material property of a substance at a given temperature. It depends on the nature of the material and the temperature,not on the dimensions of the wire such as its length or radius. Therefore,changing the radius or the length of the wire will not change the value of $Y$. The value of $Y$ can only be changed by varying the temperature of the material. Thus,none of the given options are correct.

(A) The Young's modulus $Y$ can be expressed in terms of the interatomic force constant $k$ and the interatomic distance $r$ as $Y = \frac{k}{r}$.
Given:
Interatomic distance $r = 3 \times 10^{-10} \ m$.
Interatomic force constant $k = 3.6 \times 10^{-9} \ N/\mathring{A}$.
Since $1 \mathring{A} = 10^{-10} \ m$,we have $k = 3.6 \times 10^{-9} \ N / (10^{-10} \ m) = 36 \ N/m$.
Now,substituting the values into the formula:
$Y = \frac{36 \ N/m}{3 \times 10^{-10} \ m} = 12 \times 10^{9} \ N/m^2 = 1.2 \times 10^{11} \ N/m^2$.
Therefore,the correct option is $A$.

(D) The force constant $K$ of a wire is given by the formula $K = \frac{YA}{L}$,where $Y$ is Young's modulus,$A$ is the cross-sectional area,and $L$ is the length of the wire.
Substituting $A = \pi r^2$,we get $K = \frac{Y \pi r^2}{L}$.
From this relation,it is clear that $K$ depends on:
$1$. The nature of the material (represented by $Y$).
$2$. The radius of the wire $(r)$.
$3$. The length of the wire $(L)$.
Since the force constant depends on all these factors,the correct option is $(d)$.

(D) The thermal stress developed in a wire when its expansion is prevented is given by $\sigma = Y \alpha \Delta \theta$.
Since stress $\sigma = \frac{F}{A}$,the increase in tension $\Delta F$ is given by $\Delta F = YA \alpha \Delta \theta$.
Given:
$Y = 2.2 \times 10^{11}\, N/m^2$
$A = 10^{-2}\, cm^2 = 10^{-2} \times 10^{-4}\, m^2 = 10^{-6}\, m^2$
$\alpha = 8 \times 10^{-6}\, ^\circ C^{-1}$
$\Delta \theta = 5\, ^\circ C$
Substituting these values:
$\Delta F = (2.2 \times 10^{11}) \times (10^{-6}) \times (8 \times 10^{-6}) \times 5$
$\Delta F = 2.2 \times 8 \times 5 \times 10^{11-6-6}$
$\Delta F = 88 \times 10^{-1} = 8.8\, N$.

(A) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain.
For a given material,as the density increases,the interatomic forces typically become stronger due to closer packing of atoms.
This implies that a greater force (stress) is required to produce the same amount of deformation (strain).
Since $Y = \frac{\text{Stress}}{\text{Strain}}$,an increase in the required stress for a constant strain results in an increase in the value of Young's modulus.

(C) Given: Young's modulus $Y = 10^4 \ N/m^2$,Area $A = 2 \ cm^2 = 2 \times 10^{-4} \ m^2$,Force $F = 2 \times 10^5 \ dynes = 2 \ N$.
The formula for elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$.
Substituting the values: $\Delta L = \frac{2 \times L}{(2 \times 10^{-4}) \times 10^4} = \frac{2L}{2} = L$.
The final length is the sum of the initial length and the elongation: $L_{final} = L + \Delta L = L + L = 2L$.

(B) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain.
For a solid material,$Y$ has a finite positive value.
However,for a powder,the material does not offer resistance to deformation in the same way a continuous solid does,effectively making the Young's modulus $Y = 0$.
Therefore,the correct state of the material is a solid in powder form.

(A) The formula for the extension $l$ in a wire is given by $l = \frac{FL}{AY}$,where $A = \pi r^2$ is the cross-sectional area and $Y$ is the Young's modulus.
Thus,$l = \frac{FL}{\pi r^2 Y}$.
Since the material is the same,$Y$ is constant. Therefore,$l \propto \frac{FL}{r^2}$.
For the first wire: $l_1 = l$,$F_1 = F$,$L_1 = L$,$r_1 = r$.
For the second wire: $F_2 = 2F$,$L_2 = 2L$,$r_2 = 2r$.
Taking the ratio: $\frac{l_2}{l_1} = \left( \frac{F_2}{F_1} \right) \times \left( \frac{L_2}{L_1} \right) \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{l_2}{l} = \left( \frac{2F}{F} \right) \times \left( \frac{2L}{L} \right) \times \left( \frac{r}{2r} \right)^2 = 2 \times 2 \times \left( \frac{1}{4} \right) = 1$.
Therefore,$l_2 = l$.

(D) The relationship between stress,strain,and Young's modulus $(Y)$ is given by Hooke's Law: $Y = \frac{\text{stress}}{\text{strain}}$.
Therefore,the breaking stress is calculated as: $\text{Breaking stress} = Y \times \text{strain}$.
Given: $Y = 2 \times 10^{11} \, N/m^2$ and $\text{strain} = 0.15$.
$\text{Breaking stress} = (2 \times 10^{11}) \times 0.15 = 0.30 \times 10^{11} = 3 \times 10^{10} \, N/m^2$.

(A) Given:
Area of cross-section $A = 1\,cm^2 = 1 \times 10^{-4}\,m^2$.
Young's modulus $Y = 2 \times 10^{11}\,N/m^2$.
Final length $L' = 1.1L$,so the change in length $\Delta L = L' - L = 0.1L$.
Strain $= \frac{\Delta L}{L} = 0.1$.
Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$.
Rearranging for force: $F = Y \times A \times \text{Strain}$.
$F = (2 \times 10^{11}\,N/m^2) \times (1 \times 10^{-4}\,m^2) \times (0.1)$.
$F = 2 \times 10^{11} \times 10^{-4} \times 10^{-1} = 2 \times 10^6\,N$.

(A) Elasticity is measured by the Young's modulus ($Y$ or $E$). $A$ higher value of Young's modulus indicates that the material is more elastic (i.e.,it requires more stress to produce a given strain).
The approximate values of Young's modulus for the given materials are:
$1$. Glass: $50 - 90 \, GPa$
$2$. Rubber: $0.01 - 0.1 \, GPa$
$3$. Steel: $200 \, GPa$
$4$. Copper: $117 \, GPa$
Comparing these values,Steel has the highest Young's modulus. Therefore,Steel is the most elastic substance among the given options.

(D) The stress is defined as $\text{Stress} = \frac{F}{A}$.
Since both wires are connected in series,the same force $F$ is transmitted through both wires. Given that they have the same diameter,their cross-sectional area $A$ is also the same. Therefore,the stress in both wires is the same.
The strain is defined as $\text{Strain} = \frac{\text{Stress}}{Y}$,where $Y$ is the Young's modulus of the material.
Since the Young's modulus of steel $(Y_s)$ is greater than that of copper $(Y_c)$,and the stress is the same for both,the strain in the steel wire will be less than the strain in the copper wire.
Thus,the two wires will have the same stress but different strains.

(A) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since the material $(Y)$ and the length $(L)$ are constant,we have the proportionality $l \propto \frac{F}{r^2}$.
Let the initial force be $F_1 = F$,initial radius be $r_1 = r$,and initial elongation be $l_1 = 1 \ mm$.
For the second wire,the force is $F_2 = 2F$ and the radius is $r_2 = \frac{r}{2}$.
Using the ratio: $\frac{l_2}{l_1} = \frac{F_2}{F_1} \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{l_2}{1} = \frac{2F}{F} \times \left( \frac{r}{r/2} \right)^2 = 2 \times (2)^2 = 2 \times 4 = 8$.
Therefore,the elongation $l_2 = 8 \ mm$.

(B) 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 force,we get $F = \frac{Y \cdot A \cdot \Delta l}{L}$.
Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$,we have $F \propto D^2$ (where $D$ is the diameter,and $Y, L, \Delta l$ are constant).
Given the initial force $F_1 = 10^3 \ N$ and initial diameter $D_1 = D$,the new diameter is $D_2 = 4D$.
Therefore,the ratio of forces is $\frac{F_2}{F_1} = \left(\frac{D_2}{D_1}\right)^2 = (4)^2 = 16$.
Thus,$F_2 = 16 \times F_1 = 16 \times 10^3 \ N$.

(D) Let the length of the rubber cord be $L$,cross-sectional area be $A$,and density be $d$. The mass of the cord is $M = A \cdot L \cdot d$.
When the cord is suspended vertically,the weight acts at the center of mass (at $L/2$).
The stress at a distance $x$ from the free end is due to the weight of the portion below it.
The elongation $l$ is given by the integration of strain over the length: $l = \int_{0}^{L} \frac{F(x) dx}{AY}$,where $F(x) = (A \cdot x \cdot d)g$.
Substituting $F(x)$,we get $l = \int_{0}^{L} \frac{Axdg}{AY} dx = \frac{dg}{Y} \int_{0}^{L} x dx = \frac{dg}{Y} [ \frac{x^2}{2} ]_{0}^{L} = \frac{L^2 dg}{2Y}$.
Since $g$ and $Y$ (Young's modulus) are constants,the elongation $l$ is proportional to $dL^2$.

(C) The thermal stress developed in a wire fixed at both ends when cooled is given by the formula: $F = Y A \alpha \Delta T$.
Given:
Cross-sectional area $A = 3 \, mm^2 = 3 \times 10^{-6} \, m^2$.
Young's modulus $Y = 2 \times 10^{11} \, N/m^2$.
Coefficient of linear expansion $\alpha = 10^{-5} \, ^{\circ}C^{-1}$.
Change in temperature $\Delta T = 20^{\circ}C - 10^{\circ}C = 10^{\circ}C$.
Substituting these values into the formula:
$F = (2 \times 10^{11} \, N/m^2) \times (3 \times 10^{-6} \, m^2) \times (10^{-5} \, ^{\circ}C^{-1}) \times (10^{\circ}C)$.
$F = 2 \times 3 \times 10^{11 - 6 - 5 + 1} \, N$.
$F = 6 \times 10^1 \, N = 60 \, N$.
Therefore,the tension in the wire is $60 \, N$.

(A) The elongation $l$ of a wire is given by the formula $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Since the force $F$ and Young's modulus $Y$ are constant,we have $l \propto \frac{L}{r^2}$.
Let the original length be $L_1 = L$ and radius be $r_1 = r$. The elongation is $l_1 = 0.01 \ m$.
For the second wire,the length $L_2 = 2L$ and the diameter is doubled,so the radius $r_2 = 2r$.
Using the proportionality: $\frac{l_2}{l_1} = \frac{L_2}{L_1} \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{l_2}{l_1} = \left( \frac{2L}{L} \right) \times \left( \frac{r}{2r} \right)^2 = 2 \times \left( \frac{1}{2} \right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$.
Therefore,$l_2 = \frac{l_1}{2} = \frac{0.01 \ m}{2} = 0.005 \ m$.

(A) The formula for the extension $(l)$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the applied 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 $l = \frac{FL}{\pi r^2 Y}$.
Given that the material $(Y)$,length $(L)$,and load $(F)$ are constant,the extension is inversely proportional to the square of the radius: $l \propto \frac{1}{r^2}$.
Let $l_1 = 3 \ mm$ and $r_1 = r$. Then $r_2 = \frac{r}{2}$.
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,$l_2 = 4 \times l_1 = 4 \times 3 \ mm = 12 \ mm$.

(D) The formula for extension $\Delta L$ 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.
Since $A = \pi r^2 = \pi (d/2)^2$,we have $\Delta L = \frac{4FL}{\pi d^2 Y}$.
Given that $F$ and $Y$ are constant for all wires,the extension $\Delta L \propto \frac{L}{d^2}$.
We calculate the ratio $\frac{L}{d^2}$ for each case:
$(a)$ $\frac{500}{(0.05)^2} = \frac{500}{0.0025} = 200,000$
$(b)$ $\frac{200}{(0.02)^2} = \frac{200}{0.0004} = 500,000$
$(c)$ $\frac{300}{(0.03)^2} = \frac{300}{0.0009} \approx 333,333$
$(d)$ $\frac{400}{(0.01)^2} = \frac{400}{0.0001} = 4,000,000$
Comparing the values,the ratio is maximum for case $(d)$. Therefore,the wire in case $(d)$ experiences the maximum extension.

(C) The formula for the change in length $l$ of a wire is given by $l = \frac{FL}{AY}$,where $A = \pi r^2$ is the cross-sectional area.
Since the material is the same,the Young's modulus $Y$ is constant. Given that the force $F$ is also constant,we have $l \propto \frac{L}{r^2}$.
For the first wire: $l_1 = l$,$L_1 = L$,$r_1 = r$.
For the second wire: $L_2 = 2L$,$r_2 = 2r$.
Taking the ratio: $\frac{l_2}{l_1} = \frac{L_2}{L_1} \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{l_2}{l} = \frac{2L}{L} \times \left( \frac{r}{2r} \right)^2 = 2 \times \left( \frac{1}{2} \right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$.
Therefore,the change in length of the second wire is $l_2 = \frac{l}{2}$.