Basic of Elasticity, Stress and Strain relationship and Graphical analysis Questions in English

(B) Stress is defined as the force applied per unit area: $\text{Stress} = \frac{F}{A} = \frac{F}{\pi r^2}$.
Since both wires are stretched by the same load,the force $F$ is constant.
Therefore,$\text{Stress} \propto \frac{1}{r^2}$.
Given that the radius of wire $A$ is double the radius of wire $B$,we have $r_A = 2r_B$.
Comparing the stresses $S_A$ and $S_B$:
$\frac{S_B}{S_A} = \frac{r_A^2}{r_B^2} = \left(\frac{2r_B}{r_B}\right)^2 = (2)^2 = 4$.
Thus,$S_B = 4 S_A$.
The stress on wire $B$ is four times the stress on wire $A$.

(C) The Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta l}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the elongation.
Since all wires are of the same material,$Y$ is constant. For wires of the same length $L$,we have $A = \frac{FL}{Y \Delta l}$.
For a constant load $F$,the area $A$ is inversely proportional to the elongation $\Delta l$ $(A \propto \frac{1}{\Delta l})$.
From the graph,for a given load $F$,the elongation $\Delta l$ is maximum for line $OA$ (i.e.,$\Delta l_A > \Delta l_B > \Delta l_C > \Delta l_D$).
Since $A \propto \frac{1}{\Delta l}$,the wire with the maximum elongation will have the minimum cross-sectional area.
Therefore,the line $OA$ represents the thinnest wire.

(B) Stress is defined as the force applied per unit area,given by the formula: $\text{Stress} = \frac{F}{A}$.
Given that both wires are stretched by the same load,the force $F$ is constant for both wires.
Let $A_A$ be the area of wire $A$ and $A_B$ be the area of wire $B$. According to the problem,$A_A = 2 A_B$.
The stress on wire $A$ is $\sigma_A = \frac{F}{A_A} = \frac{F}{2 A_B}$.
The stress on wire $B$ is $\sigma_B = \frac{F}{A_B}$.
Comparing the two,we get $\sigma_B = 2 \times \left(\frac{F}{2 A_B}\right) = 2 \sigma_A$.
Therefore,the stress on wire $B$ is twice that on wire $A$.

(A) The stress $\sigma$ is defined as the force $F$ divided by the cross-sectional area $A$. The elastic limit is the maximum stress the material can withstand.
Given: Elastic limit $\sigma = \frac{400}{\pi} \text{ MPa} = \frac{400}{\pi} \times 10^6 \text{ Pa}$, Force $F = 484 \text{ N}$.
The area of the cross-section of a rod with diameter $d$ is $A = \pi \frac{d^2}{4}$.
Using the formula $\sigma = \frac{F}{A}$, we have:
$\frac{400}{\pi} \times 10^6 = \frac{484}{\pi \frac{d^2}{4}}$
$\frac{400}{\pi} \times 10^6 = \frac{484 \times 4}{\pi d^2}$
Canceling $\pi$ from both sides:
$400 \times 10^6 = \frac{1936}{d^2}$
$d^2 = \frac{1936}{400 \times 10^6} = 4.84 \times 10^{-6} \text{ m}^2$
$d = \sqrt{4.84 \times 10^{-6}} = 2.2 \times 10^{-3} \text{ m} = 2.2 \text{ mm}$.

(A) Let the cross-sectional area of the rod be $a$. The area of the inclined section $AB$ is $A' = \frac{a}{\sin \theta}$.
The force $F$ acting on the rod can be resolved into two components along the section $AB$:
$1$. Normal component: $F_n = F \sin \theta$
$2$. Tangential (shearing) component: $F_s = F \cos \theta$
The shearing stress $\tau$ is defined as the tangential force divided by the area of the section:
$\tau = \frac{F_s}{A'} = \frac{F \cos \theta}{a / \sin \theta} = \frac{F \sin \theta \cos \theta}{a}$

(D) Let the natural length of the rod be $L_0$.
According to Hooke's law,the stress is proportional to strain within the elastic limit:
$\text{Stress} = Y \times \text{Strain}$
$\frac{T}{A} = Y \frac{L - L_0}{L_0}$
Rearranging for the change in length:
$L - L_0 = \frac{T L_0}{A Y}$
Let $k = \frac{L_0}{A Y}$,which is a constant for the rod.
Then,$L = L_0 + k T$.
For tension $T_1$,$L_1 = L_0 + k T_1$ --- $(i)$
For tension $T_2$,$L_2 = L_0 + k T_2$ --- (ii)
Subtracting $(i)$ from (ii):
$L_2 - L_1 = k(T_2 - T_1) \Rightarrow k = \frac{L_2 - L_1}{T_2 - T_1}$
Substitute $k$ into equation $(i)$:
$L_0 = L_1 - k T_1 = L_1 - \left( \frac{L_2 - L_1}{T_2 - T_1} \right) T_1$
$L_0 = \frac{L_1(T_2 - T_1) - T_1(L_2 - L_1)}{T_2 - T_1}$
$L_0 = \frac{L_1 T_2 - L_1 T_1 - T_1 L_2 + L_1 T_1}{T_2 - T_1}$
$L_0 = \frac{L_1 T_2 - L_2 T_1}{T_2 - T_1}$

(C) The breaking force of a wire is determined by its breaking stress,which is a material property. The breaking force $F$ is given by the product of breaking stress $\sigma$ and the cross-sectional area $A$ of the wire.
$F = \sigma \times A = \sigma \times (\pi R^2)$
Since the material is the same (copper),the breaking stress $\sigma$ remains constant.
Therefore,$F \propto R^2$.
If $F_1 = F$ for radius $R_1 = R$,and $F_2$ is the force for radius $R_2 = 2R$,then:
$\frac{F_2}{F_1} = \left(\frac{R_2}{R_1}\right)^2 = \left(\frac{2R}{R}\right)^2 = 4$
$F_2 = 4F_1 = 4F$.

(C) According to Hooke's law,within the elastic limit,stress is directly proportional to strain.
$\text{Stress} = Y \times \text{Strain}$
Where $Y$ is Young's modulus.
$\frac{F}{A} = Y \times \frac{\Delta l}{L}$
Here,$F$ is the force,$A$ is the cross-sectional area,$\Delta l$ is the change in length,and $L$ is the original length.
Rearranging for force $F$:
$F = \frac{Y A}{L} \Delta l$
Since $Y$,$A$,and $L$ are constants for a given wire,we have:
$F \propto \Delta l$
Given that the change in length is $l$,we get:
$F \propto l$
Therefore,the force is proportional to $l$.

(D) According to Hooke's Law,the extension $\Delta l$ is proportional to the applied force $T$. Let $l_0$ be the natural length and $k$ be the force constant of the string.
The length $l$ under tension $T$ is given by $l = l_0 + \frac{T}{k}$.
For $T_1 = 80 \,N$,$l_1 = 101 \,mm = l_0 + \frac{80}{k}$ --- $(1)$
For $T_2 = 100 \,N$,$l_2 = 102 \,mm = l_0 + \frac{100}{k}$ --- $(2)$
Subtracting $(1)$ from $(2)$: $102 - 101 = \frac{100-80}{k} \implies 1 = \frac{20}{k} \implies k = 20 \,N/mm$.
Substituting $k$ into $(1)$: $101 = l_0 + \frac{80}{20} \implies 101 = l_0 + 4 \implies l_0 = 97 \,mm$.
Now,for $T_3 = 160 \,N$,the length $l_3 = l_0 + \frac{T_3}{k} = 97 + \frac{160}{20} = 97 + 8 = 105 \,mm$.
Converting to cm: $105 \,mm = 10.5 \,cm$.

(C) According to Hooke's Law,the extension in a wire is proportional to the applied force: $F = k(l - l_0)$,where $l_0$ is the original length and $k$ is the force constant.
For the first case: $F_1 = k(l_1 - l_0)$ -- $(1)$
For the second case: $F_2 = k(l_2 - l_0)$ -- $(2)$
Dividing $(1)$ by $(2)$,we get: $\frac{F_1}{F_2} = \frac{l_1 - l_0}{l_2 - l_0}$
Cross-multiplying: $F_1(l_2 - l_0) = F_2(l_1 - l_0)$
$F_1 l_2 - F_1 l_0 = F_2 l_1 - F_2 l_0$
$F_2 l_0 - F_1 l_0 = F_2 l_1 - F_1 l_2$
$l_0(F_2 - F_1) = F_2 l_1 - F_1 l_2$
$l_0 = \frac{F_2 l_1 - F_1 l_2}{F_2 - F_1}$

(D) Let $L$ be the original length of the string and $k$ be the force constant of the string.
The final length is given by: $\text{Final length} = \text{Original length} + \text{Elongation}$.
Using Hooke's Law,$\text{Elongation} = \frac{F}{k}$.
Thus,$L' = L + \frac{F}{k}$.
For the first condition: $P = L + \frac{6}{k} \quad ...(i)$
For the second condition: $Q = L + \frac{8}{k} \quad ...(ii)$
To eliminate $k$,multiply equation $(i)$ by $4$ and equation $(ii)$ by $3$:
$4P = 4L + \frac{24}{k} \quad ...(iii)$
$3Q = 3L + \frac{24}{k} \quad ...(iv)$
Subtracting equation $(iv)$ from equation $(iii)$:
$4P - 3Q = (4L - 3L) + (\frac{24}{k} - \frac{24}{k})$
$4P - 3Q = L$
Therefore,the original length of the string is $4P - 3Q$.

(A) The strain is defined as the ratio of the change in length to the original length.
$\text{Strain} = \frac{\Delta \ell}{\ell}$
Given:
Original length $\ell = 40 \text{ cm}$
Change in length $\Delta \ell = 0.1 \text{ cm}$
$\text{Strain} = \frac{0.1}{40}$
$\text{Strain} = \frac{1}{400} = 0.0025$
$\text{Strain} = 25 \times 10^{-4}$

(C) According to Hooke's Law,the change in length of a wire is directly proportional to the applied force within the elastic limit.
Let $L$ be the original length and $K$ be the force constant of the wire.
For force $F_1$,the extension is $(L_1 - L)$,so $F_1 = K(L_1 - L)$ --- $(i)$
For force $F_2$,the extension is $(L_2 - L)$,so $F_2 = K(L_2 - L)$ --- (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{F_1}{F_2} = \frac{K(L_1 - L)}{K(L_2 - L)}$
$\frac{F_1}{F_2} = \frac{L_1 - L}{L_2 - L}$
Cross-multiplying gives:
$F_1(L_2 - L) = F_2(L_1 - L)$
$F_1 L_2 - F_1 L = F_2 L_1 - F_2 L$
Rearranging to solve for $L$:
$F_2 L - F_1 L = F_2 L_1 - F_1 L_2$
$L(F_2 - F_1) = F_2 L_1 - F_1 L_2$
$L = \frac{F_2 L_1 - F_1 L_2}{F_2 - F_1} = \frac{F_1 L_2 - F_2 L_1}{F_1 - F_2}$

(B) According to Hooke's Law,the extension of an elastic string is proportional to the applied tension. Let $l$ be the natural length of the string and $k$ be the force constant.
The length of the string under tension $T$ is given by $L = l + \frac{T}{k}$.
For $T_1 = 4 \ N$,$L_1 = a = l + \frac{4}{k} \implies \frac{4}{k} = a - l$ (Equation $1$).
For $T_2 = 5 \ N$,$L_2 = b = l + \frac{5}{k} \implies \frac{5}{k} = b - l$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $\frac{5}{k} - \frac{4}{k} = (b - l) - (a - l) \implies \frac{1}{k} = b - a$.
Substituting $\frac{1}{k}$ into Equation $1$: $a = l + 4(b - a) \implies a = l + 4b - 4a \implies l = 5a - 4b$.
Now,for $T_3 = 9 \ N$,the length $x$ is $x = l + \frac{9}{k}$.
Substituting $l = 5a - 4b$ and $\frac{1}{k} = b - a$:
$x = (5a - 4b) + 9(b - a) = 5a - 4b + 9b - 9a = 5b - 4a$.

(D) Let the actual length of the string be $L$ and the force constant be $k = \frac{AY}{L_0}$.
According to Hooke's Law,the extension is proportional to the applied force: $F = k \Delta L$.
For the first case: $F_1 = k(L_1 - L)$.
For the second case: $F_2 = k(L_2 - L)$.
Dividing the two equations: $\frac{F_1}{F_2} = \frac{L_1 - L}{L_2 - L}$.
Cross-multiplying gives: $F_1(L_2 - L) = F_2(L_1 - L)$.
$F_1 L_2 - F_1 L = F_2 L_1 - F_2 L$.
Rearranging to solve for $L$: $F_2 L - F_1 L = F_2 L_1 - F_1 L_2$.
$L(F_2 - F_1) = F_2 L_1 - F_1 L_2$.
Therefore,$L = \frac{F_2 L_1 - F_1 L_2}{F_2 - F_1}$.

(B) Breaking stress is defined as $\text{Stress} = \frac{F_{max}}{A}$,which implies $F_{max} = \text{Stress} \times A$.
For the lower wire:
$F_L = (12 \times 10^8 \text{ N/m}^2) \times (0.004 \times 10^{-4} \text{ m}^2) = 480 \text{ N}$.
The weight supported by the lower wire is $(m_{pan} + 10)g = 480 \text{ N}$.
$(m_{pan} + 10) \times 10 = 480 \Rightarrow m_{pan} + 10 = 48 \Rightarrow m_{pan} = 38 \text{ kg}$.
For the upper wire:
$F_U = (12 \times 10^8 \text{ N/m}^2) \times (0.008 \times 10^{-4} \text{ m}^2) = 960 \text{ N}$.
The weight supported by the upper wire is $(m_{pan} + 10 + 30)g = 960 \text{ N}$.
$(m_{pan} + 40) \times 10 = 960 \Rightarrow m_{pan} + 40 = 96 \Rightarrow m_{pan} = 56 \text{ kg}$.
To ensure both wires remain safe,we must choose the smaller mass,which is $38 \text{ kg}$.

(A) The wire must support the weight of the lift and the additional force due to acceleration. The tension $T$ in the wire is given by $T = m(g + a)$.
Given: Stress $\sigma = 4 \times 10^8 \text{ N/m}^2$,mass $m = 1600 \text{ kg}$,radius $r = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$,and $g = 10 \text{ m/s}^2$.
The cross-sectional area $A$ of the wire is $A = \pi r^2 = 3.14 \times (4 \times 10^{-3} \text{ m})^2 = 3.14 \times 16 \times 10^{-6} \text{ m}^2 = 50.24 \times 10^{-6} \text{ m}^2$.
The maximum tension $T$ the wire can withstand is $T = \sigma \times A = (4 \times 10^8 \text{ N/m}^2) \times (50.24 \times 10^{-6} \text{ m}^2) = 20096 \text{ N}$.
Using the equation of motion $T = m(g + a)$,we have $20096 = 1600(10 + a)$.
Dividing both sides by $1600$,we get $10 + a = \frac{20096}{1600} = 12.56$.
Therefore,$a = 12.56 - 10 = 2.56 \text{ m/s}^2$.