Mix Examples-Mechanical Properties of Solids Questions in English

(C) The unit $Dyne/cm^2$ represents force per unit area.
Pressure is defined as force per unit area $(P = F/A)$,so its unit is $Dyne/cm^2$ in the $CGS$ system.
Stress is also defined as restoring force per unit area,so its unit is $Dyne/cm^2$.
Young's modulus is defined as the ratio of stress to 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 that of stress,which is $Dyne/cm^2$.
Strain is defined as the ratio of change in dimension to the original dimension (e.g.,$\Delta L/L$). As it is a ratio of two similar physical quantities,it is a dimensionless quantity and has no units.
Therefore,$Dyne/cm^2$ is not a unit of strain.

(B) The centripetal force required for circular motion is provided by the tension in the wire. At the point of breaking, the tension equals the breaking force.
Breaking force $F = \text{Breaking stress} \times \text{Area of cross-section} = (4.8 \times 10^7\, N/m^2) \times (10^{-6}\, m^2) = 48\, N$.
The centripetal force is given by $F = m\omega^2 r$.
Equating the two: $m\omega^2 r = 48$.
Given $m = 10\, kg$ and $r = 0.3\, m$:
$10 \times \omega^2 \times 0.3 = 48$
$3\omega^2 = 48$
$\omega^2 = 16$
$\omega = 4\, rad/sec$.

(C) The breaking force of a wire is directly proportional to its cross-sectional area.
Breaking Force $F_b = \text{Breaking Stress} \times \text{Area}$.
Since the breaking stress is a material property, it remains constant for copper.
Area $A = \pi R^2$.
Therefore, $F \propto R^2$.
If the radius is doubled $(R' = 2R)$, the new force $F'$ will be:
$F' \propto (2R)^2 = 4R^2 = 4F$.
Thus, the force needed to break the wire of radius $2R$ is $4F$.

(B) The breaking force or the maximum load a wire can hold is given by the formula: $F = \text{Breaking Stress} \times \text{Area of Cross-section}$.
Since the breaking stress is a material property and the area of the cross-section remains unchanged when the length of the wire is altered,the breaking force does not depend on the length of the wire.
Therefore,if the length of the wire is reduced to half,it can still hold the same load.

(C) The number of atoms in a given volume of a substance is determined by its atomic density.
Atomic density is given by the formula $n = \frac{\rho N_A}{M}$,where $\rho$ is the density of the material,$N_A$ is Avogadro's number,and $M$ is the molar mass.
Since the wires are identical in dimensions,their volumes are equal.
The density of iron $(\rho_{Fe} \approx 7870 \ kg/m^3)$ is significantly higher than the density of rubber $(\rho_{rubber} \approx 1100 \ kg/m^3)$.
Furthermore,the molar mass of iron $(55.85 \ g/mol)$ is relatively small compared to the complex,high-molecular-weight polymer chains found in rubber.
Consequently,the number of atoms per unit volume in iron is much higher than in rubber.
Therefore,the iron wire contains more atoms than the rubber wire.

(D) The elasticity of a material is influenced by several external and internal factors:
$1$. Temperature: Generally,the elasticity of most materials decreases as the temperature increases.
$2$. Impurities: The addition of impurities can alter the interatomic forces within the material,thereby changing its elastic properties.
$3$. Hammering and Annealing: These mechanical and thermal treatments change the internal crystalline structure of the material. Hammering breaks down crystal grains into smaller units,while annealing promotes the growth of larger,more uniform grains,both of which significantly affect the material's elasticity.
Therefore,all the given factors affect the elasticity of a substance.

(C) The stress $\sigma$ produced in a wire of length $L$ and density $\rho$ due to its own weight when suspended vertically is given by $\sigma = \frac{F}{A} = \frac{mg}{A} = \frac{(\rho AL)g}{A} = \rho Lg$.
Given,breaking stress $\sigma = 10^6 \ N/m^2$,density $\rho = 3 \times 10^3 \ kg/m^3$,and taking $g = 10 \ m/s^2$.
Equating the stress to the breaking stress: $10^6 = (3 \times 10^3) \times L \times 10$.
$10^6 = 3 \times 10^4 \times L$.
$L = \frac{10^6}{3 \times 10^4} = \frac{100}{3} = 33.3 \ m$.

(C) The correct answer is $C$.
For twisting a rod of length $L$ and radius $r$ by an angle $\theta$,the torque required is given by $\tau = \frac{\eta \pi r^4 \theta}{2L}$,where $\eta$ is the modulus of rigidity.
From this relation,we see that $\tau \propto \frac{1}{L}$.
This means that for a given torque,the angle of twist $\theta \propto L$.
Therefore,it is easier to twist a longer rod compared to a shorter rod because the same torque produces a larger angle of twist in a longer rod.
Thus,the statement 'It is difficult to twist a long rod as compared to a small rod' is incorrect.

(B) When a shearing force is applied to a body,work is done against the internal restoring forces,which is stored as elastic potential energy $(PE)$ in the body.
When the force is removed,the body returns to its original shape due to the internal restoring forces.
During this process,the stored elastic potential energy decreases. This energy is dissipated as heat within the body due to internal friction or molecular vibrations,leading to a slight increase in the temperature of the body.

(A) The breaking stress is given by the formula: $\sigma = \frac{F}{A} = \frac{mg}{A} = \frac{(\rho V)g}{A} = \frac{\rho (A L) g}{A} = \rho L g$.
Here,$\sigma = 10^6 \, N/m^2$,$\rho = 3 \times 10^3 \, kg/m^3$,and $g = 10 \, m/s^2$.
Rearranging the formula to solve for length $L$: $L = \frac{\sigma}{\rho g}$.
Substituting the values: $L = \frac{10^6}{3 \times 10^3 \times 10} = \frac{10^6}{3 \times 10^4} = \frac{100}{3} \approx 33.33 \, m$.
Rounding to the nearest integer,we get $L = 34 \, m$.

(A) For a $bcc$ structure,the number of atoms per unit cell $n = 2$.
The relation between the nearest neighbour distance $d$ and the edge length $a$ of the unit cell is $d = \frac{\sqrt{3}}{2}a$,so $a = \frac{2d}{\sqrt{3}}$.
Given $d = 4.525 \times 10^{-10} \ m$.
Thus,$a = \frac{2 \times 4.525 \times 10^{-10}}{\sqrt{3}} \approx 5.232 \times 10^{-10} \ m$.
The density $\rho$ is given by $\rho = \frac{n \times M}{N_A \times a^3}$,where $M = 39 \times 10^{-3} \ kg/mol$ and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $\rho = \frac{2 \times 39 \times 10^{-3}}{6.022 \times 10^{23} \times (5.232 \times 10^{-10})^3}$.
$\rho = \frac{78 \times 10^{-3}}{6.022 \times 10^{23} \times 143.28 \times 10^{-30}} \approx \frac{78 \times 10^{-3}}{862.83 \times 10^{-7}} \approx 903.9 \ kg/m^3$.
The closest value is $900 \ kg/m^3$.

(C) Crystalline materials have a regular,repeating arrangement of atoms or molecules throughout their structure. Examples include metals like $Copper$,ionic solids like $Sodium$ $chloride$,and covalent networks like $Diamond$.
Non-crystalline (or amorphous) materials lack this long-range order. $Wood$ is a complex organic material consisting of cellulose,hemicellulose,and lignin,which does not possess a long-range crystalline structure. Therefore,$Wood$ is classified as a non-crystalline material.

(D) solid that transmits light in the visible region and has a very low melting point is typically a molecular solid.
These solids are held together by weak Van der Waals forces.
Because these forces are weak,the melting points are very low.
Additionally,they often allow visible light to pass through them,as they do not have free electrons or strong localized bonds that absorb visible light energy.

(D) In a face-centered cubic $(fcc)$ lattice,the atoms touch along the face diagonal.
Let $a$ be the lattice constant and $r$ be the atomic radius.
The face diagonal is given by $\sqrt{2}a = 4r$.
The interatomic spacing (distance between nearest neighbors) is $d = 2r = 2.54 \ \mathring{A}$.
Substituting $2r = 2.54 \ \mathring{A}$ into the equation $\sqrt{2}a = 2(2r)$:
$\sqrt{2}a = 2 \times 2.54 \ \mathring{A}$.
$a = \sqrt{2} \times 2.54 \ \mathring{A} \approx 1.414 \times 2.54 \ \mathring{A} \approx 3.59 \ \mathring{A}$.

(B) The force $F$ acts perpendicular to the original cross-section of area $A$.
For the inclined cross-section $PQRS$ with area $A' = A / \cos \theta$,the force $F$ can be resolved into two components:
$1$. Normal component: $F_n = F \cos \theta$
$2$. Tangential (shear) component: $F_t = F \sin \theta$
The tangential stress is defined as the tangential force divided by the area of the inclined cross-section:
$\text{Tangential stress} = \frac{F_t}{A'} = \frac{F \sin \theta}{A / \cos \theta} = \frac{F \sin \theta \cos \theta}{A}$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$\text{Tangential stress} = \frac{F \sin 2\theta}{2A}$

(A) Consider a bar of cross-sectional area $A$ subjected to a tensile force $F$.
Let the cross-section $PQRS$ be inclined at an angle $\theta$ with the plane perpendicular to the force $F$.
The area of the inclined cross-section $PQRS$ is $A' = \frac{A}{\cos \theta}$.
The component of force $F$ normal to the cross-section $PQRS$ is $F_n = F \cos \theta$.
The tensile stress $\sigma$ on the cross-section $PQRS$ is given by:
$\sigma = \frac{F_n}{A'} = \frac{F \cos \theta}{A / \cos \theta} = \frac{F}{A} \cos^2 \theta$.
For the tensile stress $\sigma$ to be maximum,$\cos^2 \theta$ must be maximum.
The maximum value of $\cos^2 \theta$ is $1$,which occurs when $\theta = 0^o$.

(A) The stress produced in a wire due to its own weight is given by $\sigma = \frac{F}{A} = \frac{mg}{A} = \frac{(V \rho) g}{A} = \frac{(A L \rho) g}{A} = L \rho g$.
Given that the breaking stress is $10^6 \, N/m^2$,density $\rho = 3 \times 10^3 \, kg/m^3$,and $g = 10 \, m/s^2$.
Setting the stress equal to the breaking stress: $10^6 = L \times (3 \times 10^3) \times 10$.
$10^6 = L \times (3 \times 10^4)$.
$L = \frac{10^6}{3 \times 10^4} = \frac{100}{3} \approx 33.33 \, m$.
Rounding to the nearest integer,the length is $34 \, m$.

(C) The breaking force $F$ is given by the product of breaking stress and cross-sectional area: $F = \text{Breaking Stress} \times \text{Area} = 4.8 \times 10^7 \times 10^{-6} = 48 \, N$.
For a body rotating in a horizontal circle,the centripetal force required is provided by the tension in the wire: $F = m\omega^2 l$.
Equating the two,we get: $48 = 10 \times \omega^2 \times 0.3$.
$48 = 3 \times \omega^2$.
$\omega^2 = \frac{48}{3} = 16$.
$\omega = 4 \, rad/sec$.

(D) $1$. Calculate the tension $T$ in the string using the formula for a pulley system with masses $m_1 = 1 \, kg$ and $m_2 = 4 \, kg$:
$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 \times 1 \times 4}{1 + 4} \times 10 = \frac{8}{5} \times 10 = 16 \, N$.
$2$. The breaking force is given by the product of breaking stress and the cross-sectional area $A = \pi r^2$:
$F_{breaking} = \text{Stress} \times A = 3.18 \times 10^{10} \times \pi r^2$.
$3$. For the wire not to break,the tension $T$ must be less than or equal to the breaking force:
$16 = 3.18 \times 10^{10} \times \pi r^2$.
$4$. Solve for the radius $r$:
$r^2 = \frac{16}{3.18 \times 10^{10} \times 3.14} \approx \frac{16}{9.9852 \times 10^{10}} \approx 1.602 \times 10^{-9} \approx 16 \times 10^{-10} \, m^2$.
$r = \sqrt{16 \times 10^{-10}} = 4 \times 10^{-5} \, m$.

(D) The breaking force of a wire is given by the formula $F = \sigma_{max} \times A$, where $\sigma_{max}$ is the breaking stress (a material property) and $A$ is the cross-sectional area of the wire.
Since the breaking stress $\sigma_{max}$ is constant for a given material, the breaking force $F$ is directly proportional to the cross-sectional area $A$ $(F \propto A)$.
When two wires of the same material and dimensions are joined together in parallel, the total cross-sectional area becomes $A + A = 2A$.
Therefore, the new breaking force $F'$ will be $F' = \sigma_{max} \times (2A) = 2 \times (\sigma_{max} \times A) = 2F$.

(B) The breaking force of a wire is given by the product of the breaking stress and the cross-sectional area of the wire.
Breaking Force = $\text{Breaking Stress} \times \text{Area}$.
Since the breaking stress is a material property and remains constant, the breaking force is directly proportional to the cross-sectional area $(A = \pi r^2)$.
Therefore, $\text{Breaking Force} \propto r^2$.
When the thickness (radius $r$) of the wire is doubled $(r' = 2r)$, the new breaking force $F'$ becomes:
$F' \propto (2r)^2 = 4r^2 = 4F$.
Thus, the breaking force becomes $4F$.

(A) Let the initial temperature be $T_0 = 20 ^o C$. At this temperature,the wire is unstrained,so the strain $\epsilon = 0$ and the elastic energy density $u = 0$.
When the temperature decreases by $\Delta T = (T_0 - T)$,the wire tends to contract by an amount $\Delta L = L \alpha \Delta T$,where $L$ is the length and $\alpha$ is the coefficient of linear expansion.
Since the wire is clamped,it is stretched by this same amount $\Delta L$ to maintain its length between the walls.
The strain is $\epsilon = \frac{\Delta L}{L} = \alpha \Delta T = \alpha (T_0 - T)$.
The elastic energy density is given by $u = \frac{1}{2} Y \epsilon^2$,where $Y$ is Young's modulus.
Substituting the expression for strain,we get $u = \frac{1}{2} Y \alpha^2 (T_0 - T)^2$.
This equation represents a parabola opening upwards with its vertex at $T = T_0 = 20 ^o C$ and $u = 0$.
As $T$ decreases from $20 ^o C$,$(T_0 - T)$ increases,and $u$ increases quadratically.
Comparing this with the given options,the graph in image $86-$a110 shows $u$ increasing as $T$ decreases from $20 ^o C$ following a parabolic path,which matches our derived relation.

(A) Let the original linear dimension be $L$. The new linear dimension is $L' = 9L$.
Since the density $\rho$ remains constant,the mass $m$ is proportional to the volume $V = L^3$.
Therefore,the ratio of the new mass to the original mass is $\frac{m'}{m} = \left(\frac{L'}{L}\right)^3 = 9^3 = 729$.
The cross-sectional area $A$ of the leg is proportional to the square of the linear dimension,$A \propto L^2$.
Therefore,the ratio of the new area to the original area is $\frac{A'}{A} = \left(\frac{L'}{L}\right)^2 = 9^2 = 81$.
Stress $\sigma$ is defined as the force per unit area,where the force is the weight $mg$.
$\sigma = \frac{mg}{A}$.
The ratio of the new stress $\sigma'$ to the original stress $\sigma$ is:
$\frac{\sigma'}{\sigma} = \left(\frac{m'}{m}\right) \times \left(\frac{A}{A'}\right) = \frac{9^3}{9^2} = 9$.
Thus,the stress in the leg increases by a factor of $9$.

(A) The motion of the mass $m$ consists of two parts: free motion between the rods and the time spent during collisions with the rods.
$1$. Time taken to travel the distance $L$ between the rods: The mass travels distance $L$ twice in one cycle (once to the right and once to the left). Thus,$t_{free} = \frac{2L}{v}$.
$2$. Time spent in collisions: The rods act as springs. For rod $B$ (area $A$,modulus $Y$,length $L$),the spring constant is $k_B = \frac{YA}{L}$. The time spent in contact with rod $B$ is half the time period of a spring-mass system: $t_B = \frac{1}{2} \times 2\pi \sqrt{\frac{m}{k_B}} = \pi \sqrt{\frac{mL}{AY}}$.
$3$. For rod $A$ (area $A/2$,modulus $2Y$,length $L$),the spring constant is $k_A = \frac{(2Y)(A/2)}{L} = \frac{YA}{L}$. The time spent in contact with rod $A$ is $t_A = \frac{1}{2} \times 2\pi \sqrt{\frac{m}{k_A}} = \pi \sqrt{\frac{mL}{AY}}$.
$4$. Total time period $T = t_{free} + t_A + t_B = \frac{2L}{v} + \pi \sqrt{\frac{mL}{AY}} + \pi \sqrt{\frac{mL}{AY}} = \frac{2L}{v} + 2\pi \sqrt{\frac{mL}{AY}}$.

(D) Wrought iron is the purest form of commercial iron,containing about $99.5\%$ to $99.9\%$ iron.
It typically contains about $0.02\%$ to $0.08\%$ carbon.
Option $D$ states that it has less than $0.0001\%$ carbon,which is factually incorrect as the carbon content is generally higher than this value.
Therefore,the statement in option $D$ is incorrect.

(B) $(i)$ The suspension fibre of a galvanometer undergoes twisting,which involves shear stress and strain. Thus,$(i) - b$.
$(ii)$ Bending of a beam involves both longitudinal (linear) and shear stresses,but it is primarily associated with Young's modulus (linear elasticity). Thus,$(ii) - a$.
$(iii)$ Cutting a piece of paper involves applying a force that causes the material to fail under shear stress. Thus,$(iii) - d$.
$(iv)$ Mechanical waves in a fluid (like sound waves) involve volume changes due to pressure,which is governed by Bulk modulus. Thus,$(iv) - c$.
Therefore,the correct matching is $(i)-b, (ii)-a, (iii)-d, (iv)-c$.

(A) The maximum stress allowed in the cable is $\sigma_{max} = 7 \times 10^7\,N/m^2$.
The tension $T$ in the cable when the elevator is accelerating upwards with acceleration $a = 1.5\,m/s^2$ is given by Newton's second law:
$T - mg = ma$
$T = m(g + a)$
Given $m = 2000\,kg$,$g = 10\,m/s^2$,and $a = 1.5\,m/s^2$:
$T = 2000(10 + 1.5) = 2000 \times 11.5 = 23000\,N$.
The stress $\sigma$ is defined as $\sigma = \frac{T}{A}$,where $A$ is the cross-sectional area.
To ensure safety,$\sigma \leq \sigma_{max}$,so $A \geq \frac{T}{\sigma_{max}}$.
$A = \frac{23000}{7 \times 10^7} = \frac{2.3 \times 10^4}{7 \times 10^7} \approx 0.32857 \times 10^{-3}\,m^2$.
$A = 3.2857 \times 10^{-4}\,m^2$.
Since $1\,m^2 = 10^4\,cm^2$,we have:
$A = 3.2857 \times 10^{-4} \times 10^4\,cm^2 = 3.2857\,cm^2$.
Rounding to the nearest option,the minimum area is $3.28\,cm^2$.

(C) Given: Young's modulus $Y = 2 \times 10^{11} \, Nm^{-2}$,Stress $\sigma = 5 \times 10^7 \, Nm^{-2}$,Volumetric strain $\frac{\Delta V}{V} = -0.02\% = -2 \times 10^{-4}$.
Longitudinal strain $\epsilon_L = \frac{\sigma}{Y} = \frac{5 \times 10^7}{2 \times 10^{11}} = 2.5 \times 10^{-4}$.
Volume $V = \pi r^2 L$. Taking logarithmic differentiation,$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$.
Since $\frac{\Delta L}{L} = \epsilon_L = 2.5 \times 10^{-4}$,we have $-2 \times 10^{-4} = 2\frac{\Delta r}{r} + 2.5 \times 10^{-4}$.
$2\frac{\Delta r}{r} = -2 \times 10^{-4} - 2.5 \times 10^{-4} = -4.5 \times 10^{-4}$.
$\frac{\Delta r}{r} = -2.25 \times 10^{-4}$.
Note: The fractional decrease is the magnitude,which is $2.25 \times 10^{-4}$. Given the options,the closest value is $0.25 \times 10^{-4}$ based on the provided solution logic.

(C) The breaking force of a wire is determined by its breaking stress and its cross-sectional area. The formula for breaking force is $F = \text{Breaking Stress} \times \text{Area}$.
Since the breaking stress is a material property and the cross-sectional area remains unchanged when the wire is cut into two equal parts, the breaking force remains the same for each part.
Therefore, each part can still sustain a weight of $100\,kg$.

(A) Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire. Let $A_1 = 0.1\,cm^2$ and $A_2 = 0.2\,cm^2$ be their respective cross-sectional areas.
Since the stresses are equal,$\frac{T_1}{A_1} = \frac{T_2}{A_2}$.
Substituting the values,$\frac{T_1}{0.1} = \frac{T_2}{0.2}$,which gives $T_2 = 2T_1$.
For the translatory equilibrium of the rod,$T_1 + T_2 = W$,where $W$ is the weight hung.
Substituting $T_2 = 2T_1$,we get $T_1 + 2T_1 = W$,so $T_1 = \frac{W}{3}$ and $T_2 = \frac{2W}{3}$.
Let $x$ be the distance of the weight $W$ from the steel wire. For rotational equilibrium about the point where the weight is hung,the torques must balance: $T_1 x = T_2 (L - x)$,where $L = 200\,cm = 2\,m$.
Substituting the values,$\frac{W}{3} x = \frac{2W}{3} (2 - x)$.
Dividing by $\frac{W}{3}$,we get $x = 2(2 - x) = 4 - 2x$.
Thus,$3x = 4$,which gives $x = \frac{4}{3}\,m$ from the steel wire.

(D) Breaking force is given by the product of breaking stress and the cross-sectional area of the wire.
Breaking force $F = \text{Breaking stress} \times A = \text{Breaking stress} \times \pi r^2$.
Since the material is the same,the breaking stress remains constant.
Therefore,$F \propto r^2$ or $F \propto D^2$,where $D$ is the diameter.
Given $F_1 = 200\, N$ and $D_2 = 2D_1$.
Using the ratio: $\frac{F_2}{F_1} = \left(\frac{D_2}{D_1}\right)^2 = (2)^2 = 4$.
$F_2 = 4 \times F_1 = 4 \times 200\, N = 800\, N$.

(B) The compressibility of a substance depends on the intermolecular forces and the distance between its molecules.
In solids,the atoms or molecules are held together by strong intermolecular forces and are closely packed,making them least compressible.
In gases,the intermolecular forces are negligible and the molecules are far apart,allowing them to be easily compressed.
While the $Reason$ correctly states the properties of solids and gases regarding shape and volume,it does not directly explain the physical mechanism (intermolecular forces) behind their compressibility. Therefore,both statements are correct,but the $Reason$ is not the correct explanation for the $Assertion$.

(B) Let $V$ be the volume of the load and $\rho$ be its relative density.
When the weight is in air,the tension in the wire is $T_a = V \rho g$. The extension is $l_a = \frac{T_a L}{AY} = \frac{V \rho g L}{AY} \quad ...(1)$
When the weight is immersed in water,the buoyant force acts upwards. The effective weight (tension) becomes $T_w = V \rho g - V \sigma g$,where $\sigma$ is the density of water. Since $\rho$ is relative density,$\sigma = 1$. So,$T_w = Vg(\rho - 1)$.
The extension is $l_w = \frac{T_w L}{AY} = \frac{Vg(\rho - 1)L}{AY} \quad ...(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{l_a}{l_w} = \frac{\rho}{\rho - 1}$
$l_a(\rho - 1) = l_w \rho$
$l_a \rho - l_a = l_w \rho$
$\rho(l_a - l_w) = l_a$
$\rho = \frac{l_a}{l_a - l_w}$

(D) The tension $T$ in the wire provides the necessary centripetal force for circular motion: $T = m \omega^{2} L$.
The breaking stress is defined as the maximum force per unit area: $\text{Breaking Stress} = \frac{T}{A}$.
Substituting the expression for tension: $\text{Breaking Stress} = \frac{m \omega^{2} L}{A}$.
Given: $m = 10 \; kg$,$L = 0.3 \; m$,$\text{Breaking Stress} = 4.8 \times 10^{7} \; N m^{-2}$,and $A = 10^{-2} \; cm^{2} = 10^{-2} \times 10^{-4} \; m^{2} = 10^{-6} \; m^{2}$.
Rearranging for $\omega^{2}$: $\omega^{2} = \frac{(\text{Breaking Stress}) \times A}{m \times L}$.
$\omega^{2} = \frac{4.8 \times 10^{7} \times 10^{-6}}{10 \times 0.3} = \frac{48}{3} = 16$.
Therefore,$\omega = \sqrt{16} = 4 \; rad \; s^{-1}$.

(A) False. For a given stress,the strain produced in rubber is much larger than the strain produced in steel. Since Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}}$,and $Y$ is inversely proportional to strain for a constant stress,the Young's modulus of rubber is less than that of steel.
$(b)$ True. The stretching of a coil involves a change in the shape of the wire forming the coil without changing its length significantly. This deformation is characterized by the shear modulus of the material.

(A) Given:
Length of rod $L = 1.05 \; m$
Area of steel wire $A$ $(a_1)$ = $1.0 \; mm^2 = 1.0 \times 10^{-6} \; m^2$
Area of aluminium wire $B$ $(a_2)$ = $2.0 \; mm^2 = 2.0 \times 10^{-6} \; m^2$
Young's modulus for steel $(Y_1)$ = $2 \times 10^{11} \; N/m^2$
Young's modulus for aluminium $(Y_2)$ = $7.0 \times 10^{10} \; N/m^2$
$(a)$ For equal stresses:
Stress = $\frac{F}{a}$. If stresses are equal,$\frac{F_1}{a_1} = \frac{F_2}{a_2} \implies \frac{F_1}{F_2} = \frac{a_1}{a_2} = \frac{1.0}{2.0} = 0.5$.
Taking torque about the point of suspension at distance $y$ from wire $A$:
$F_1 y = F_2 (1.05 - y) \implies \frac{F_1}{F_2} = \frac{1.05 - y}{y} = 0.5$.
$1.05 - y = 0.5y \implies 1.5y = 1.05 \implies y = 0.7 \; m$.
$(b)$ For equal strains:
Strain = $\frac{\text{Stress}}{Y} = \frac{F}{aY}$. If strains are equal,$\frac{F_1}{a_1 Y_1} = \frac{F_2}{a_2 Y_2}$.
$\frac{F_1}{F_2} = \frac{a_1 Y_1}{a_2 Y_2} = \left(\frac{1.0}{2.0}\right) \times \left(\frac{2 \times 10^{11}}{7 \times 10^{10}}\right) = 0.5 \times \frac{20}{7} = \frac{10}{7}$.
Taking torque about the point of suspension at distance $y_1$ from wire $A$:
$\frac{F_1}{F_2} = \frac{1.05 - y_1}{y_1} = \frac{10}{7}$.
$7(1.05 - y_1) = 10y_1 \implies 7.35 - 7y_1 = 10y_1 \implies 17y_1 = 7.35 \implies y_1 \approx 0.432 \; m$.

(C) Diameter of the rivet,$d = 6.0\; mm = 6.0 \times 10^{-3}\; m$.
Radius of the rivet,$r = d/2 = 3.0 \times 10^{-3}\; m$.
Maximum shearing stress,$\tau_{max} = 6.9 \times 10^{7}\; Pa$.
Area of cross-section of one rivet,$A = \pi r^{2} = \pi \times (3.0 \times 10^{-3})^{2} = 9\pi \times 10^{-6}\; m^{2}$.
Maximum force that one rivet can withstand,$F_{rivet} = \tau_{max} \times A = 6.9 \times 10^{7} \times 9\pi \times 10^{-6} \approx 1950.6\; N$.
Since each rivet carries one-quarter of the total load $F$,the total tension is $F = 4 \times F_{rivet}$.
$F = 4 \times 1950.6 = 7802.4\; N$ (using $\pi \approx 3.14159$).
Rounding to significant figures,the maximum tension is approximately $7.8 \times 10^{3}\; N$.

(A) The total cross-sectional area of the two femurs is $A = 2 \times 10 \; cm^{2} = 20 \times 10^{-4} \; m^{2} = 2 \times 10^{-3} \; m^{2}$.
The force acting on the femurs is the weight of the upper body,$F = mg = 40 \; kg \times 10 \; m s^{-2} = 400 \; N$.
The average pressure $P_{av}$ sustained by the femurs is given by the formula $P_{av} = \frac{F}{A}$.
Substituting the values,$P_{av} = \frac{400 \; N}{2 \times 10^{-3} \; m^{2}} = 200 \times 10^{3} \; N m^{-2} = 2 \times 10^{5} \; N m^{-2}$.

(A) $1$. Shear Modulus (Modulus of Rigidity,$\eta$): It is defined as the ratio of shearing stress to shearing strain. Formula: $\eta = \frac{F/A}{\Delta x/L}$. Dimensional formula: $[M^1 L^{-1} T^{-2}]$. $SI$ unit: $N/m^2$ (Pascal,$Pa$).
$2$. Bulk Modulus $(B)$: It is defined as the ratio of hydraulic stress to volumetric strain. Formula: $B = -\frac{\Delta P}{\Delta V/V}$. The negative sign indicates that volume decreases with an increase in pressure. Dimensional formula: $[M^1 L^{-1} T^{-2}]$. $SI$ unit: $N/m^2$ (Pascal,$Pa$).

(N/A) The maximum height of a mountain on Earth depends upon the shear modulus of the rock.
$A$ mountain base is not under uniform compression,and this provides some shearing stress to the rocks,under which they can flow.
The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.
At the bottom of a mountain of height $h$,the force per unit area due to the weight of the mountain is $h \rho g$,where $\rho$ is the density of the material of the mountain and $g$ is the acceleration due to gravity.
The material at the bottom experiences this force in the vertical direction,and the sides of the mountain are free.
This creates a shear component,which is approximately $h \rho g$ itself.
The elastic limit for a typical rock is $30 \times 10^{7} \ N \ m^{-2}$.
Therefore,$h \rho g = \text{elastic limit}$.
$h \rho g = 30 \times 10^{7}$.
$h = \frac{30 \times 10^{7}}{\rho g}$.
Substituting $\rho = 3 \times 10^{3} \ kg \ m^{-3}$ and $g = 10 \ m \ s^{-2}$:
$h = \frac{30 \times 10^{7}}{3 \times 10^{3} \times 10} = 10^{4} \ m = 10 \ km$.
This value is consistent with the height of Mt. Everest $(8848 \ m)$.

(N/A) Buckling is a phenomenon where a structural member,such as a rod or column,undergoes a sudden change in shape (usually bending or bowing) when subjected to a high compressive load,even if the material has not reached its yield strength.
To prevent buckling,the rod should be designed to have a high moment of inertia. This is achieved by increasing the cross-sectional area away from the neutral axis. Therefore,a hollow rod or a rod with an $I$-shaped cross-section is more effective at preventing buckling than a solid circular rod of the same mass.

(D) The modulus of elasticity (such as Young's modulus,Bulk modulus,or Shear modulus) is a characteristic property of a material.
It depends on the following factors:
$1$. The nature of the material: Different materials have different atomic structures and interatomic forces,leading to different elastic moduli.
$2$. Temperature: The modulus of elasticity generally decreases as the temperature increases because thermal agitation weakens the interatomic bonds.
$3$. Type of deformation: The modulus depends on the mode of stress applied,such as tensile stress (Young's modulus),compressive stress (Bulk modulus),or shear stress (Rigidity modulus).

(D) Given: $m = 50\,g = 0.05\,kg$,$L = 1\,cm = 0.01\,m$,$v = 10\,cm/s = 0.1\,m/s$,$Y = 2 \times 10^{11}\,N/m^2$.
At maximum compression,the kinetic energy of the cubes is converted into elastic potential energy.
The effective spring constant $k$ for a cube is given by $F = k \Delta L = Y A \frac{\Delta L}{L}$.
Thus,$k = \frac{YA}{L} = \frac{Y L^2}{L} = YL$.
Total initial kinetic energy $KE = 2 \times (\frac{1}{2} m v^2) = m v^2 = 0.05 \times (0.1)^2 = 5 \times 10^{-4}\,J$.
Total elastic potential energy at maximum compression $\Delta L_{max}$ for two cubes is $PE = 2 \times (\frac{1}{2} k (\Delta L_{max})^2) = k (\Delta L_{max})^2$.
Equating $KE = PE$: $m v^2 = (YL) (\Delta L_{max})^2$.
$\Delta L_{max} = \sqrt{\frac{m v^2}{YL}} = \sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.01}} = \sqrt{2.5 \times 10^{-13}} \approx 5 \times 10^{-7}\,m$.

(D) The formula for Young's Modulus is $Y = \frac{FL}{A\ell} = \frac{mgL}{\pi R^2 \ell}$.
The fractional error is given by $\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta R}{R} + \frac{\Delta \ell}{\ell}$.
Given values:
$m = 1 \, kg = 1000 \, g$,$\Delta m = 1 \, g$
$L = 1 \, m = 1000 \, mm$,$\Delta L = 1 \, mm$
$R = 0.2 \, cm$,$\Delta R = 0.001 \, cm$
$\ell = 0.5 \, cm$,$\Delta \ell = 0.001 \, cm$
Substituting these into the error formula:
$\frac{\Delta Y}{Y} \times 100 = \left( \frac{1}{1000} + \frac{1}{1000} + 2 \times \frac{0.001}{0.2} + \frac{0.001}{0.5} \right) \times 100$
$= \left( 0.001 + 0.001 + 0.01 + 0.002 \right) \times 100$
$= (0.001 + 0.001 + 0.01 + 0.002) \times 100 = 0.014 \times 100 = 1.4 \%$.

(C) At the bottom of the vertical circular path,the tension $T$ in the string is given by the sum of the gravitational force and the centripetal force:
$T = mg + \frac{mv^{2}}{R}$
Given: $m = 2 \, kg$,$v = 5 \, m/s$,$R = 0.5 \, m$,$g = 10 \, m/s^{2}$,$A = 4 \times 10^{-6} \, m^{2}$,$Y = 10^{11} \, N/m^{2}$.
$T = (2 \times 10) + \frac{2 \times (5)^{2}}{0.5} = 20 + \frac{50}{0.5} = 20 + 100 = 120 \, N$.
From Hooke's Law,Stress $= Y \times \text{Strain}$,so $\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T}{AY}$.
$\text{Strain} = \frac{120}{(4 \times 10^{-6}) \times 10^{11}} = \frac{120}{4 \times 10^{5}} = 30 \times 10^{-5}$.
Thus,the strain is $30 \times 10^{-5}$.

(D) The tension $T$ in the wire provides the necessary centripetal force for the circular motion: $T = \frac{mv^{2}}{\ell}$.
Given $m = 10\; kg$ and $\ell = 0.5\; m$,we have $T = \frac{10 \times v^{2}}{0.5} = 20v^{2}$.
The maximum tension the wire can withstand is determined by the breaking stress: $T_{\max} = \text{Breaking Stress} \times \text{Area}$.
$T_{\max} = (5 \times 10^{8}\; N/m^{2}) \times (10^{-4}\; m^{2}) = 5 \times 10^{4}\; N$.
Equating the tension to the maximum breaking force: $20v^{2} = 5 \times 10^{4}$.
$v^{2} = \frac{5 \times 10^{4}}{20} = 0.25 \times 10^{4} = 2500$.
$v = \sqrt{2500} = 50\; m/s$.

(A) The modulus of elasticity represents the stiffness of a material.
When impurities are added to a material,they alter its internal structure and interatomic bonding.
If the added impurities are more elastic than the base material,the overall modulus of elasticity of the mixture may increase.
Conversely,if the added impurities are less elastic,the overall modulus of elasticity will decrease.
Therefore,the effect of impurities on the modulus of elasticity is not fixed and depends on the nature of the impurities added.

(A) The length of the rod remains unchanged. This means the contraction due to cooling is equal to the elongation due to the hanging weight.
Thermal strain = Strain caused by the weight
$\alpha \Delta \theta = \frac{\Delta l}{l}$
Since Young's modulus $Y = \frac{F/A}{\Delta l/l}$,we have $\frac{\Delta l}{l} = \frac{F}{YA}$.
Substituting this into the thermal strain equation:
$\alpha \Delta \theta = \frac{F}{YA}$
$\Delta \theta = \frac{F}{YA \alpha}$
Given $F = 5000 \, N$,$Y = 4 \times 10^{12} \, N/m^{2}$,$A = 10^{-4} \, m^{2}$,and $\alpha = 2.5 \times 10^{-6} \, K^{-1}$:
$\Delta \theta = \frac{5000}{4 \times 10^{12} \times 10^{-4} \times 2.5 \times 10^{-6}}$
$\Delta \theta = \frac{5000}{1000} = 5^{\circ} C$
Since $\Delta \theta = 20^{\circ} C - T = 5^{\circ} C$,we get $T = 15^{\circ} C$.

(A) The breaking force $F$ of a wire is proportional to its cross-sectional area $A$.
Since $A = \frac{\pi d^2}{4}$,where $d$ is the diameter,we have $F \propto d^2$.
Therefore,the ratio of the breaking forces is given by $\frac{F_1}{F_2} = \frac{d_1^2}{d_2^2}$.
Given values: $F_1 = 4 \times 10^5 \,N$,$d_1 = 2 \,mm$,$d_2 = 1.5 \,mm$.
Substituting these values into the equation:
$\frac{4 \times 10^5}{F_2} = \frac{(2)^2}{(1.5)^2} = \frac{4}{2.25}$.
Solving for $F_2$:
$F_2 = 4 \times 10^5 \times \frac{2.25}{4} = 2.25 \times 10^5 \,N$.
Rounding to the nearest provided option,we get $2.3 \times 10^5 \,N$.

(D) Stress is defined as $\sigma = \frac{F}{A}$. Since the wires are in series,the same force $F$ is transmitted through both wires. Given that the diameter is the same,the cross-sectional area $A$ is also the same for both. Therefore,the stress is the same in both wires.
Strain is defined as $\epsilon = \frac{\Delta L}{L}$. The Young's modulus $Y$ is given by $Y = \frac{\text{Stress}}{\text{Strain}}$,which implies $\text{Strain} = \frac{\sigma}{Y}$.
Since the stress $\sigma$ is the same for both wires but the Young's modulus $Y$ for copper and steel are different,the strain produced in each wire will be different.