Rigidity Modulus Questions in English

(D) The modulus of rigidity (also known as shear modulus) is defined as the ratio of shear stress to shear strain.
$\text{Modulus of rigidity} = \frac{\text{Shear Stress}}{\text{Shear Strain}}$
Since shear strain is a dimensionless quantity (it is a ratio of two lengths),the unit of the modulus of rigidity is the same as the unit of stress.
Stress is defined as force per unit area,so its unit is $\frac{N}{m^2}$ or $\text{Pascal} (Pa)$.
Therefore,the correct unit is $N/m^2$.

(D) The modulus of rigidity $(\eta)$ is defined as the ratio of shear stress to shear strain.
Shear stress is defined as force per unit area,so its dimensional formula is $[F]/[A] = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$.
Shear strain is a dimensionless quantity because it is the ratio of two lengths.
Therefore,the dimensional formula for the modulus of rigidity is the same as that of stress,which is $[ML^{-1}T^{-2}]$.
Thus,the correct option is $D$.

(C) Shear modulus is defined as the ratio of shearing stress to shearing strain.
Shearing stress = $\frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Shearing strain is a dimensionless quantity.
Therefore,the dimensions of shear modulus are the same as those of stress,which is $[ML^{-1}T^{-2}]$.

(B) The modulus of rigidity (also known as shear modulus) measures the resistance of a material to shear deformation. Diamond is the hardest known natural material,characterized by a very strong covalent crystal lattice structure. Due to this extremely strong atomic bonding,it exhibits the highest resistance to deformation among all known materials. Therefore,its modulus of rigidity is greater than that of any other material. The correct option is $B$.

(D) The modulus of rigidity (also known as the shear modulus) is an intrinsic property of the material of the rod.
Since both rods $A$ and $B$ are made of the same material,their modulus of rigidity will be identical.
Therefore,the ratio of the modulus of rigidity of $A$ to $B$ is $1:1$.

(C) The twisting couple $C$ required to produce an angle of twist $\theta$ in a wire of length $l$,radius $r$,and modulus of rigidity $\eta$ is given by the formula:
$C = \frac{\pi \eta r^4 \theta}{2l}$
From this expression,we can see that for a constant twisting couple $C$,length $l$,and material property $\eta$:
$\theta = \frac{2lC}{\pi \eta r^4}$
This implies that $\theta \propto \frac{1}{r^4}$.
Therefore,for wires $A$ and $B$ with radii $r_1$ and $r_2$ respectively,the ratio of the angles of twist $\theta_1$ and $\theta_2$ is:
$\frac{\theta_1}{\theta_2} = \frac{r_2^4}{r_1^4}$
Thus,the correct option is $C$.

(C) The modulus of rigidity,also known as the shear modulus,is defined as the ratio of shear stress to shear strain.
Liquids do not have a fixed shape and offer no resistance to shear stress.
When a shear stress is applied to a liquid,it begins to flow continuously,resulting in an infinite shear strain for any finite shear stress.
Since the modulus of rigidity is defined as $\text{Shear Stress} / \text{Shear Strain}$,and the shear strain becomes infinitely large for any applied shear stress,the modulus of rigidity for a liquid is $0$.

(B) The angle of shear $\phi$ is given by the formula $\phi = \frac{r\theta}{L}$,where $r$ is the radius of the wire,$\theta$ is the angle of twist in radians,and $L$ is the length of the wire.
Given: $r = 4\, mm = 0.4\, cm$,$L = 100\, cm$,and $\theta = 30^\circ$.
Substituting the values into the formula:
$\phi = \frac{0.4\, cm}{100\, cm} \times 30^\circ$
$\phi = 0.004 \times 30^\circ = 0.12^\circ$.
Therefore,the angle of shear is $0.12^\circ$.

(D) The torsional rigidity of a rod is given by $C = \frac{\pi \eta r^4}{2L}$,where $\eta$ is the modulus of rigidity.
Since the rods are joined in series,the torque $\tau$ transmitted through both rods is the same.
Let $\theta_0$ be the twist angle at the joint.
For the small rod (length $l/2$,radius $r/2$): $\tau = C_1(\theta_0 - 0) = \frac{\pi \eta (r/2)^4}{2(l/2)} \theta_0 = \frac{\pi \eta r^4}{16(l/2)} \theta_0 = \frac{\pi \eta r^4}{8l} \theta_0$.
For the large rod (length $l$,radius $r$): $\tau = C_2(\theta - \theta_0) = \frac{\pi \eta r^4}{2l} (\theta - \theta_0)$.
Equating the torques: $\frac{\pi \eta r^4}{8l} \theta_0 = \frac{\pi \eta r^4}{2l} (\theta - \theta_0)$.
$\frac{\theta_0}{8} = \frac{\theta - \theta_0}{2} \Rightarrow \theta_0 = 4(\theta - \theta_0) \Rightarrow \theta_0 = 4\theta - 4\theta_0 \Rightarrow 5\theta_0 = 4\theta \Rightarrow \theta_0 = 4\theta/5$.
Wait,re-evaluating the expression: $\frac{\theta_0}{4} = \theta - \theta_0 \Rightarrow \theta_0 = 4\theta - 4\theta_0 \Rightarrow 5\theta_0 = 4\theta$.
Let's re-check the torsional constant: $C = \frac{\pi \eta r^4}{2L}$.
$C_1 = \frac{\pi \eta (r/2)^4}{2(l/2)} = \frac{\pi \eta r^4 / 16}{l} = \frac{\pi \eta r^4}{16l}$.
$C_2 = \frac{\pi \eta r^4}{2l}$.
$C_1 \theta_0 = C_2(\theta - \theta_0) \Rightarrow \frac{\pi \eta r^4}{16l} \theta_0 = \frac{\pi \eta r^4}{2l} (\theta - \theta_0) \Rightarrow \frac{\theta_0}{8} = \theta - \theta_0 \Rightarrow \theta_0 = 8\theta - 8\theta_0 \Rightarrow 9\theta_0 = 8\theta \Rightarrow \theta_0 = 8\theta/9$.

(C) The torque $\tau$ required to produce an angular displacement $\theta$ in a wire of length $l$,radius $r$,and modulus of rigidity $\eta$ is given by $\tau = \frac{\pi \eta r^4 \theta}{2l}$.
Since the length $l$ and material (modulus of rigidity $\eta$) are the same for both wires,and the applied torque $\tau$ is equal,we have:
$\tau_1 = \tau_2$
$\frac{\pi \eta r_1^4 \theta_1}{2l} = \frac{\pi \eta r_2^4 \theta_2}{2l}$
$r_1^4 \theta_1 = r_2^4 \theta_2$
$\frac{\theta_1}{\theta_2} = \frac{r_2^4}{r_1^4}$

(C) The modulus of rigidity $\eta$ is defined as the ratio of shearing stress to shearing strain:
$\eta = \frac{F/A}{x/L}$
where $F$ is the tangential force,$A$ is the area of the face on which the force is applied,$x$ is the lateral displacement,and $L$ is the height of the block perpendicular to the force.
Rearranging for displacement $x$:
$x = \frac{F \cdot L}{\eta \cdot A}$
Since $\eta$ and $F$ are constant for all cases,the displacement $x$ is proportional to the ratio of height to area: $x \propto \frac{L}{A}$.
Calculating $L/A$ for each case:
For position $P$: $L = 5\,cm$,$A = 10 \times 8 = 80\,cm^2$. So,$L/A = 5/80 = 0.0625\,cm^{-1}$.
For position $Q$: $L = 10\,cm$,$A = 8 \times 5 = 40\,cm^2$. So,$L/A = 10/40 = 0.25\,cm^{-1}$.
For position $R$: $L = 8\,cm$,$A = 10 \times 5 = 50\,cm^2$. So,$L/A = 8/50 = 0.16\,cm^{-1}$.
Comparing the values,the ratio $L/A$ is maximum in position $Q$. Therefore,the displacement is maximum in position $Q$.

(C) The shearing stress is defined as the ratio of the shearing force to the area of the surface parallel to the force.
Given,shearing strength (maximum shearing stress) $\tau_{max} = 345 \, MN/m^2 = 345 \times 10^6 \, N/m^2$.
The area of the face $ABCD$ subjected to shear is $A = BC \times BE = 2 \, cm \times 2 \, cm = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$.
The maximum shearing force $F$ that the rod can withstand is $F = \tau_{max} \times A$.
$F = (345 \times 10^6 \, N/m^2) \times (4 \times 10^{-4} \, m^2) = 345 \times 4 \times 10^2 \, N = 138000 \, N$.
Since $F = mg$,where $m$ is the mass of the load and $g = 10 \, m/s^2$:
$m = F / g = 138000 / 10 = 13800 \, kg$.
Thus,the maximum load is $13800 \, kg$.

(A) The block is fixed to an inclined plane. Gravity acts on the block,which can be resolved into two components: one perpendicular to the inclined plane $(mg \cos \theta)$ and one parallel to the inclined plane $(mg \sin \theta)$.
The component $mg \sin \theta$ acts as a shear force on the block,causing it to deform due to its low shear modulus. This deformation is a shear strain,which results in the block tilting in the direction of the force.
Since the base is fixed,the top surface of the block will shift parallel to the inclined plane while the vertical sides will tilt. This results in the block taking the shape of a parallelogram,as shown in option $A$.

(B) For the same material,the modulus of rigidity $\eta$ is constant,where $\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}}$.
Shear stress is given by $\sigma = \frac{F}{A}$,where $A = L^2$ is the area of the face.
Shear strain is given by $\gamma = \frac{\Delta x}{L}$,where $\Delta x$ is the displacement and $L$ is the side length.
Thus,$\eta = \frac{F/L^2}{\Delta x/L} = \frac{F}{L \cdot \Delta x}$.
For the first cube: $L_1 = 10\,cm = 0.1\,m$,$\Delta x_1 = 0.5\,cm = 0.005\,m$,$F = 10^5\,N$.
For the second cube: $L_2 = 20\,cm = 0.2\,m$,$\Delta x_2 = x$,$F = 10^5\,N$.
Since the material is the same,$\eta_1 = \eta_2$:
$\frac{F}{L_1 \cdot \Delta x_1} = \frac{F}{L_2 \cdot \Delta x_2}$
$L_1 \cdot \Delta x_1 = L_2 \cdot \Delta x_2$
$10\,cm \times 0.5\,cm = 20\,cm \times x$
$5 = 20x$
$x = \frac{5}{20} = 0.25\,cm$.

(C) To punch a hole of diameter $D = 1 \, cm = 10^{-2} \, m$ in a steel sheet of thickness $h = 0.3 \, cm = 0.3 \times 10^{-2} \, m$,the shearing force must act along the cylindrical surface area of the hole.
The shearing stress $\sigma_{max}$ is given as $3.5 \times 10^8 \, N \, m^{-2}$.
The area $A$ resisting the shear is the lateral surface area of the cylinder being punched out:
$A = \text{Circumference} \times \text{thickness} = (\pi D) \times h$
Substituting the values:
$A = \pi \times (10^{-2} \, m) \times (0.3 \times 10^{-2} \, m) = 0.3 \pi \times 10^{-4} \, m^2$
The force $F$ required is:
$F = \sigma_{max} \times A$
$F = (3.5 \times 10^8 \, N \, m^{-2}) \times (0.3 \pi \times 10^{-4} \, m^2)$
$F = 3.5 \times 0.3 \times 3.14159 \times 10^4 \, N$
$F \approx 3.298 \times 10^4 \, N$
Rounding to the nearest value,we get $F \approx 3.3 \times 10^4 \, N$.

(A) The shearing force acts on the area of the boundary (perimeter) of the hole.
The perimeter of the square hole is $P = 4 \times \text{side} = 4 \times 2\,cm = 8\,cm = 0.08\,m$.
The thickness of the sheet is $t = 2\,mm = 0.002\,m = 2 \times 10^{-3}\,m$.
The area $A$ over which the shearing force acts is the product of the perimeter and the thickness:
$A = P \times t = 0.08\,m \times 0.002\,m = 1.6 \times 10^{-4}\,m^2$.
Given the shearing stress $\tau = 3.5 \times 10^8\,N/m^2$, the force $F$ is calculated as:
$F = \tau \times A$
$F = (3.5 \times 10^8\,N/m^2) \times (1.6 \times 10^{-4}\,m^2)$
$F = 5.6 \times 10^4\,N$.

(C) Given:
Length of wire $L = 1\,m = 100\,cm$.
Radius of wire $r = 4\,mm = 0.4\,cm$.
Angle of twist $\theta = 30^\circ$.
From the geometry of a twisted cylinder,the arc length $BB^{\prime}$ can be expressed in two ways:
$BB^{\prime} = r \theta = L \phi$
where $\phi$ is the angle of shear.
Rearranging for $\phi$:
$\phi = \frac{r \theta}{L}$
Substituting the values:
$\phi = \frac{0.4\,cm \times 30^\circ}{100\,cm}$
$\phi = \frac{12}{100} = 0.12^\circ$.
Thus,the angle of shear is $0.12^\circ$.

(D) Given: Radius $r = 4 \, mm = 4 \times 10^{-3} \, m$,Length $L = 100 \, cm = 1 \, m$,and angle of twist $\theta = 60^o = 60 \times \frac{\pi}{180} \, rad = \frac{\pi}{3} \, rad$.
For a wire twisted by an angle $\theta$,the angle of shear $\phi$ at the surface is given by the relation $\phi = \frac{r \theta}{L}$.
Substituting the values: $\phi = \frac{4 \times 10^{-3} \, m \times (60^o)}{1 \, m} = 4 \times 10^{-3} \times 60^o = 0.24^o$.

(D) Shear modulus,also known as the modulus of rigidity,measures the resistance of a material to shear deformation.
It is defined only for solids because they possess a definite shape and can resist tangential forces.
Fluids (liquids and gases) do not have a fixed shape and cannot sustain shear stress; they flow when subjected to such forces.
Therefore,the shear modulus for both liquids and gases is $0$.

(N/A) The lead slab is fixed at the bottom,and a shearing force $F = 9.0 \times 10^{4} \, N$ is applied parallel to the top narrow face.
The area $A$ of the face on which the force is applied is:
$A = 50 \, cm \times 10 \, cm = 0.5 \, m \times 0.1 \, m = 0.05 \, m^2$
The shearing stress is given by:
$\text{Stress} = \frac{F}{A} = \frac{9.0 \times 10^{4} \, N}{0.05 \, m^2} = 1.8 \times 10^{6} \, N/m^2$
We know that the shearing strain is defined as:
$\text{Strain} = \frac{\Delta x}{L} = \frac{\text{Stress}}{G}$
Where $L = 50 \, cm = 0.5 \, m$ is the height of the slab and $G = 5.6 \times 10^{9} \, N/m^2$ is the shear modulus of lead.
Therefore,the displacement $\Delta x$ is:
$\Delta x = \frac{\text{Stress} \times L}{G} = \frac{1.8 \times 10^{6} \, N/m^2 \times 0.5 \, m}{5.6 \times 10^{9} \, N/m^2}$
$\Delta x = \frac{0.9 \times 10^{6}}{5.6 \times 10^{9}} \, m \approx 0.1607 \times 10^{-3} \, m$
$\Delta x \approx 1.6 \times 10^{-4} \, m = 0.16 \, mm$

(C) The shear modulus $\eta$ is defined as the ratio of shear stress to shear strain:
$\eta = \frac{F/A}{\Delta L/L}$
Where $F$ is the force applied,$A$ is the area of the face,$L$ is the side length of the cube,and $\Delta L$ is the vertical deflection.
Rearranging for $\Delta L$:
$\Delta L = \frac{F L}{A \eta}$
Given:
$F = mg = 100\; kg \times 9.8\; m/s^2 = 980\; N$
$L = 10\; cm = 0.1\; m$
$A = L^2 = (0.1\; m)^2 = 0.01\; m^2 = 10^{-2}\; m^2$
$\eta = 25\; GPa = 25 \times 10^9\; Pa$
Substituting the values:
$\Delta L = \frac{980 \times 0.1}{10^{-2} \times 25 \times 10^9}$
$\Delta L = \frac{98}{0.25 \times 10^9} = 392 \times 10^{-9}\; m = 3.92 \times 10^{-7}\; m$
The vertical deflection is $3.92 \times 10^{-7}\; m$.

(B) The shear modulus $(G)$ is defined as the ratio of shear stress to shear strain.
Liquids do not have a fixed shape and cannot sustain a tangential (shear) force.
When a tangential force is applied to a liquid,it flows continuously,meaning the shear strain increases indefinitely over time.
Since the liquid cannot resist shear stress,the shear modulus for a liquid is $0$.

(B) When a spring is stretched,the wire of the spring undergoes a change in shape (twisting/shearing) rather than just a change in length. Therefore,the stiffness of the spring is determined by the shear modulus $(G)$ of the material.
Assertion $(A)$ is true.
Regarding Reason $(R)$,steel has a much higher Young's modulus and tensile strength compared to copper. Therefore,a steel spring is stronger and more resistant to deformation than a copper spring of the same dimensions. Thus,the statement that a copper spring has more tensile strength than a steel spring is false.
Reason $(R)$ is false.
Therefore,the correct option is $(B)$.

(C) The shear modulus $\eta$ is given by the formula: $\eta = \frac{F/A}{x/L}$,where $F$ is the shearing force,$A$ is the area of the face on which the force is applied,$x$ is the displacement,and $L$ is the side length of the slab.
Rearranging for displacement $x$: $x = \frac{F \cdot L}{A \cdot \eta}$.
Given: $F = 18.0 \times 10^{4} \, N$,$L = 60 \, cm = 0.6 \, m$,thickness $t = 15 \, cm = 0.15 \, m$,and $\eta = 25 \times 10^{9} \, N m^{-2}$.
The area $A$ of the narrow face is $L \times t = 0.6 \, m \times 0.15 \, m = 0.09 \, m^2$.
Substituting the values: $x = \frac{18.0 \times 10^{4} \times 0.6}{0.09 \times 25 \times 10^{9}}$.
$x = \frac{10.8 \times 10^{4}}{2.25 \times 10^{9}} = 4.8 \times 10^{-5} \, m = 48 \times 10^{-6} \, m = 48 \, \mu m$.

(B) The modulus of rigidity $(G)$ is defined as the ratio of shear stress to shear strain.
$G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}$
Given:
$F = 10 \, kN = 10 \times 10^3 \, N$
$L = 4 \, cm = 0.04 \, m$
$A = 1 \, cm^2 = 1 \times 10^{-4} \, m^2$
$G = 8 \times 10^{11} \, N/m^2$
Rearranging the formula to solve for lateral displacement $x$:
$x = \frac{FL}{AG}$
Substituting the values:
$x = \frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times 8 \times 10^{11}}$
$x = \frac{400}{8 \times 10^7} = 50 \times 10^{-7} = 5 \times 10^{-6} \, m$
Thus,the lateral displacement is $5 \times 10^{-6} \, m$.

(B) The twisting couple (torque) $\tau$ applied to a wire is given by the formula: $\tau = \frac{\pi \eta r^4 \phi}{2L}$,where $\eta$ is the modulus of rigidity,$r$ is the radius,$\phi$ is the angle of twist,and $L$ is the length of the wire.
Since the wires have the same length $L$,same material (same $\eta$),and are subjected to the same twisting couple $\tau$,we have: $\tau = \frac{\pi \eta r_1^4 \phi_A}{2L} = \frac{\pi \eta r_2^4 \phi_B}{2L}$.
Equating the two expressions: $r_1^4 \phi_A = r_2^4 \phi_B$.
Therefore,the ratio of the angle of twist at the end of $A$ to the angle of twist at the end of $B$ is: $\frac{\phi_A}{\phi_B} = \frac{r_2^4}{r_1^4}$.

(D) The shear modulus $\eta$ is defined as the ratio of shear stress to shear strain $\phi$:
$\eta = \frac{\text{Shear stress}}{\phi}$
Therefore,the shear stress is given by:
$\text{Shear stress} = \eta \phi$
The strain energy per unit volume is given by the formula:
$\text{Energy density} = \frac{1}{2} \times \text{Shear stress} \times \text{Shear strain}$
Substituting the values:
$\frac{\text{Strain energy}}{V} = \frac{1}{2} \times (\eta \phi) \times \phi$
$\frac{\text{Strain energy}}{V} = \frac{1}{2} \eta \phi^2$
Multiplying by volume $V$,we get the total strain energy:
$\text{Strain energy} = \frac{1}{2} \eta \phi^2 V$
Thus,the correct option is $D$.

(C) The elastic potential energy $U$ stored in a twisted wire is equal to the work done in twisting it.
The formula for the work done in twisting a wire of length $L$,radius $r$,and modulus of rigidity $\eta$ by an angle $\alpha$ is given by:
$U = \frac{1}{2} C \alpha^2$
where $C$ is the torsional rigidity (or torque constant) of the wire.
The torsional rigidity $C$ is given by:
$C = \frac{\pi \eta r^4}{2 L}$
Substituting the value of $C$ into the energy equation:
$U = \frac{1}{2} \left( \frac{\pi \eta r^4}{2 L} \right) \alpha^2$
$U = \frac{\pi \eta r^4 \alpha^2}{4 L}$
Thus,the correct option is $C$.

(B) Given: Diameter $d = 12\,mm$,so radius $r = 6\,mm = 6 \times 10^{-3}\,m$.
Length of the wire $\ell = 1\,m$.
Angle of twist $\theta = 30^{\circ}$.
The relationship between the angle of shear $\phi$ and the angle of twist $\theta$ for a wire is given by $r \theta = \ell \phi$.
Therefore,$\phi = \frac{r \theta}{\ell}$.
Substituting the values: $\phi = \frac{6 \times 10^{-3}\,m \times 30^{\circ}}{1\,m} = 180 \times 10^{-3}\,^{\circ} = 0.18^{\circ}$.
Thus,the angle of shear is $0.18^{\circ}$.

(B) The shear stress $\sigma$ is defined as the force $F$ applied per unit area $A$. For a slab of side $l$ and thickness $d$,the area of the face on which the force is applied is $A = l \times d$.
Thus,the shear stress for the two slabs is:
$\sigma_1 = \frac{F}{l d_1}$ and $\sigma_2 = \frac{F}{l d_2}$.
The shear modulus $\eta$ is defined as the ratio of shear stress to the angle of deformation $\theta$ (for small angles,$\theta \approx \tan \theta$):
$\eta = \frac{\sigma}{\theta} \Rightarrow \theta = \frac{\sigma}{\eta}$.
Given $\theta_2 = 2\theta_1$,we substitute the expressions:
$\frac{\sigma_2}{\eta_2} = 2 \left( \frac{\sigma_1}{\eta_1} \right)$.
Substituting $\sigma_1 = \frac{F}{l d_1}$,$\sigma_2 = \frac{F}{l d_2}$,and $d_2 = 2d_1$:
$\frac{F}{l d_2 \eta_2} = 2 \left( \frac{F}{l d_1 \eta_1} \right)$.
$\frac{1}{2d_1 \eta_2} = \frac{2}{d_1 \eta_1} \Rightarrow \frac{1}{\eta_2} = \frac{4}{\eta_1} \Rightarrow \eta_2 = \frac{\eta_1}{4}$.
Given $\eta_1 = 4 \times 10^9 \ N/m^2$,we get:
$\eta_2 = \frac{4 \times 10^9}{4} = 1 \times 10^9 \ N/m^2$.
Comparing this with $x \times 10^9 \ N/m^2$,we find $x = 1$.

(A) The shear modulus $G$ is defined as the ratio of shear stress to shear strain (angular displacement $\theta$ for small angles).
$G = \frac{\text{Shear Stress}}{\theta}$
Shear stress $\sigma = \frac{F}{A}$,where $F = 10^5 \ N$ and $A = \pi r^2$.
Given $r = 4 \ cm = 0.04 \ m = 4 \times 10^{-2} \ m$.
Area $A = \pi \times (4 \times 10^{-2})^2 = 16 \pi \times 10^{-4} \ m^2$.
Now,substitute the values into the formula:
$10^{10} = \frac{10^5 / (16 \pi \times 10^{-4})}{\theta}$
$\theta = \frac{10^5}{16 \pi \times 10^{-4} \times 10^{10}}$
$\theta = \frac{10^5}{16 \pi \times 10^6} = \frac{1}{160 \pi} \text{ radian}$.

(B) Given: Length of the rod $L = 2 \ m = 200 \ cm$. Radius of the rod $R = 1 \ cm$. Angle of twist $\theta = 0.8 \ rad$.
For a rod twisted at one end,the relationship between the shear strain $\phi$,the radius $R$,the angle of twist $\theta$,and the length $L$ is given by the formula: $L \phi = R \theta$.
Substituting the values into the equation:
$\phi = \frac{R \theta}{L} = \frac{1 \ cm \times 0.8 \ rad}{200 \ cm}$.
$\phi = \frac{0.8}{200} = 0.004 \ rad$.
Therefore,the shear strain developed is $0.004 \ rad$.

(B) The shear modulus $\eta$ is defined as $\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\Delta x/L}$,where $A$ is the area of the face,$\Delta x$ is the displacement,and $L$ is the side length.
For the first cube: $\eta = \frac{F/L^2}{x_{1}/L} = \frac{F}{L x_{1}}$.
Thus,$x_{1} = \frac{F}{\eta L}$.
For the second cube of side $2L$: The area $A' = (2L)^2 = 4L^2$. The shear modulus $\eta$ remains the same as the material is the same.
Let the new displacement be $x_{2}$. Then $\eta = \frac{F/A'}{x_{2}/(2L)} = \frac{F/(4L^2)}{x_{2}/(2L)} = \frac{F}{2L x_{2}}$.
Equating the two expressions for $\eta$: $\frac{F}{L x_{1}} = \frac{F}{2L x_{2}}$.
Solving for $x_{2}$: $x_{2} = \frac{x_{1}}{2}$.

(A) Let $l$ be the length of the wire,$r$ be the radius,$\theta$ be the angle of twist,and $\phi$ be the angle of shear.
From the geometry of the twisted wire,the arc length $s$ on the circumference is given by $s = r \theta$.
For small angles,the angle of shear $\phi$ is related to the arc length $s$ and the length of the wire $l$ by $\phi = \frac{s}{l}$.
Substituting $s = r \theta$,we get $\phi = \frac{r \theta}{l}$.
Given: $l = 1 \text{ m}$,$r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$,and $\theta = 45^{\circ}$.
Substituting these values: $\phi = \frac{2 \times 10^{-3} \text{ m} \times 45^{\circ}}{1 \text{ m}} = 90 \times 10^{-3} \text{ degrees} = 0.09^{\circ}$.
Therefore,the angle of shear is $0.09^{\circ}$.

(A) Given: Force $F = 50 \text{ kN} = 50 \times 10^3 \text{ N}$.
Dimensions: $40 \text{ mm} \times 20 \text{ mm} = 40 \times 10^{-3} \text{ m} \times 20 \times 10^{-3} \text{ m} = 800 \times 10^{-6} \text{ m}^2$.
Shear Modulus $G = 40 \times 10^9 \text{ Nm}^{-2}$.
Stress $\sigma = \frac{F}{A} = \frac{50 \times 10^3}{800 \times 10^{-6}} = \frac{50 \times 10^9}{800} = 0.0625 \times 10^9 \text{ Nm}^{-2} = 6.25 \times 10^7 \text{ Nm}^{-2}$.
Strain $\epsilon = \frac{\text{Stress}}{G} = \frac{6.25 \times 10^7}{40 \times 10^9} = \frac{6.25}{40} \times 10^{-2} = 0.15625 \times 10^{-2} = 1.5625 \times 10^{-3}$.
Thus,the strain is approximately $1.56 \times 10^{-3}$.

(B) When a load is suspended from a spiral spring,the wire of the spring experiences a twisting effect.
This twisting effect creates a torque that leads to a change in the shape of the wire cross-section without changing its volume.
This type of deformation is characterized by shearing stress and shearing strain.
Therefore,the strain produced in the wire of a spiral spring when stretched is shearing strain.

(C) The modulus of rigidity $\eta$ is defined as the ratio of tangential stress to shear strain: $\eta = \frac{F/A}{\Delta x/L}$,where $F$ is the tangential force,$A$ is the area of the face,$\Delta x$ is the lateral displacement,and $L$ is the side length of the cube.
Given: $L = 5 \ cm = 0.05 \ m$,$A = L^2 = (0.05 \ m)^2 = 25 \times 10^{-4} \ m^2$,$F = 1800 \ N$,and $\eta = 2.4 \times 10^6 \ N \ m^{-2}$.
Rearranging the formula for lateral displacement $\Delta x$: $\Delta x = \frac{F \cdot L}{A \cdot \eta}$.
Substituting the values: $\Delta x = \frac{1800 \times 0.05}{25 \times 10^{-4} \times 2.4 \times 10^6}$.
$\Delta x = \frac{90}{6000} = 0.015 \ m$.
Converting to millimeters: $\Delta x = 0.015 \times 1000 \ mm = 15 \ mm$.

(B) The vertical deflection $x$ of a cantilever rod due to a load $F = mg$ at its end is related to the shear modulus $\eta$ by the formula:
$\eta = \frac{F L}{A x} \Rightarrow x = \frac{mg L}{A \eta}$
Where $L = 6 \text{ cm} = 6 \times 10^{-2} \text{ m}$,radius $r = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$,$m = 400 \pi \text{ kg}$,$g = 10 \text{ m/s}^2$,and $\eta = 3 \times 10^{10} \text{ N/m}^2$.
The cross-sectional area $A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4 \pi \times 10^{-4} \text{ m}^2$.
Substituting these values into the formula:
$x = \frac{(400 \pi \times 10) \times (6 \times 10^{-2})}{(4 \pi \times 10^{-4}) \times (3 \times 10^{10})}$
$x = \frac{4000 \pi \times 6 \times 10^{-2}}{12 \pi \times 10^6}$
$x = \frac{24000 \pi \times 10^{-2}}{12 \pi \times 10^6} = \frac{240 \pi}{12 \pi \times 10^6} = 20 \times 10^{-6} \text{ m}$
$x = 20 \times 10^{-3} \text{ mm} = 0.02 \text{ mm}$.
Thus,the correct option is $B$.

(B) Given: Side of the slab $L = 50 \text{ cm} = 0.5 \text{ m}$, thickness $t = 10 \text{ cm} = 0.1 \text{ m}$, shearing force $F = 10^5 \text{ N}$, displacement $x = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$.
Area of the face subjected to force $A = L \times t = 0.5 \text{ m} \times 0.1 \text{ m} = 0.05 \text{ m}^2$.
Shear stress $= \frac{F}{A} = \frac{10^5}{0.05} = 2 \times 10^6 \text{ N/m}^2$.
Shear strain $= \frac{x}{L} = \frac{0.2 \times 10^{-3} \text{ m}}{0.5 \text{ m}} = 0.4 \times 10^{-3} = 4 \times 10^{-4}$.
Shear modulus $\eta = \frac{\text{Shear stress}}{\text{Shear strain}} = \frac{2 \times 10^6}{4 \times 10^{-4}} = 0.5 \times 10^{10} \text{ N/m}^2 = 5 \times 10^9 \text{ N/m}^2 = 5 \text{ GPa}$.

(B) The modulus of rigidity $\eta$ is defined as the ratio of shear stress to shear strain: $\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{x/L}$.
Rearranging the formula to solve for displacement $x$: $x = \frac{FL}{A\eta}$.
Given values: side length $L = 5$ cm $= 0.05$ m,area $A = L^2 = (0.05)^2 = 0.0025$ m$^2$,force $F = 10$ $N$,and modulus of rigidity $\eta = 10^5$ $N$/m$^2$.
Substituting these values into the formula: $x = \frac{10 \times 0.05}{0.0025 \times 10^5}$.
$x = \frac{0.5}{250} = 0.002$ m.
Converting meters to millimeters: $0.002$ m $= 2$ mm.