### Related Formula By Hooke's Law, the tension (T$T$) in a stretched elastic string is proportional to its change in length: T = k cdot Delta L = k(L - L_0)$T = k \cdot \Delta L = k(L - L_0)$ where: k$k$ = spring constant of the string L$L$ = current stretched length L_0$L_0$ = natural original length ### Core Logic We can set up two algebraic equations using the given conditions: 1. **Case 1:** Tension T_1 = 5 \ mathrmN$T_1 = 5 \ \mathrm{N}$ produces stretched length L_1 = 1.4 \ mathrmm$L_1 = 1.4 \ \mathrm{m}$: 5 = k(1.4 - L_0) quad dots (1)$5 = k(1.4 - L_0) \quad \dots (1)$ 2. **Case 2:** Tension T_2 = 7 \ mathrmN$T_2 = 7 \ \mathrm{N}$ produces stretched length L_2 = 1.56 \ mathrmm$L_2 = 1.56 \ \mathrm{m}$: 7 = k(1.56 - L_0) quad dots (2)$7 = k(1.56 - L_0) \quad \dots (2)$ ### Step 1: Solve for natural length Divide equation (1) by equation (2) to eliminate the spring constant k$k$: frac57 = frac1.4 - L_01.56 - L_0$\frac{5}{7} = \frac{1.4 - L_0}{1.56 - L_0}$ Cross-multiply and solve for L_0$L_0$: 5(1.56 - L_0) = 7(1.4 - L_0)$5(1.56 - L_0) = 7(1.4 - L_0)$ 7.8 - 5 L_0 = 9.8 - 7 L_0$7.8 - 5 L_0 = 9.8 - 7 L_0$ Rearrange to isolate the variable: 2 L_0 = 9.8 - 7.8 = 2$2 L_0 = 9.8 - 7.8 = 2$ L_0 = 1 \ mathrmm$L_0 = 1 \ \mathrm{m}$ Thus, the original length of the string is 1 mathrm~m$1 \mathrm{~m}$. ### Pattern Recognition Sees: Elastic stretching of a string/wire under variable load. Trap: Attempting to resolve details like Young's modulus or cross-sectional area. Taking ratios allows the spring constant k$k$ to cancel cleanly, saving computational effort. Shortcut: Use the ratio equation fracT_1T_2 = fracL_1 - L_0L_2 - L_0$\frac{T_1}{T_2} = \frac{L_1 - L_0}{L_2 - L_0}$ directly. Solving the linear equation yields L_0 = 1text m$L_0 = 1\text{ m}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula sigma = fracepsilon_texttransverseepsilon_textlongitudinal$\sigma = \frac{\epsilon_{\text{transverse}}}{\epsilon_{\text{longitudinal}}}$ u = frac12 Y epsilon_textlongitudinal^2$u = \frac{1}{2} Y \epsilon_{\text{longitudinal}}^2$ ### Core Logic Given parameters: - Young's modulus, Y = 2.0 times 10^11mathrm~N/m^2$Y = 2.0 \times 10^{11}\mathrm{~N/m}^2$ - Poisson's ratio, sigma = 0.2$\sigma = 0.2$ - Transverse strain, epsilon_texttrans = 10^-3$\epsilon_{\text{trans}} = 10^{-3}$ First, calculate the longitudinal strain epsilon$\epsilon$: sigma = fracepsilon_texttransepsilon implies 0.2 = frac10^-3epsilon$\sigma = \frac{\epsilon_{\text{trans}}}{\epsilon} \implies 0.2 = \frac{10^{-3}}{\epsilon}$ epsilon = frac10^-30.2 = 5 times 10^-3$\epsilon = \frac{10^{-3}}{0.2} = 5 \times 10^{-3}$ Now, compute the elastic potential energy density u$u$: u = frac12 Y epsilon^2 = frac12 times (2.0 times 10^11) times (5 times 10^-3)^2$u = \frac{1}{2} Y \epsilon^2 = \frac{1}{2} \times (2.0 \times 10^{11}) \times (5 \times 10^{-3})^2$ u = 10^11 times 25 times 10^-6 = 25 times 10^5mathrm~J/m^3$u = 10^{11} \times 25 \times 10^{-6} = 25 \times 10^5\mathrm{~J/m}^3$ Expressing as x times 10^5$x \times 10^5$, we have x = 25$x = 25$. ### Step 1: Final Conclusion The value of the coefficient is 25. ### Pattern Recognition Poisson's ratio is defined as the ratio of lateral/transverse strain to longitudinal strain. Use this to determine the longitudinal stretch, then apply the basic potential energy density formula frac12 Y epsilon^2$\frac{1}{2} Y \epsilon^2$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula Young's modulus Y$Y$ is defined as: Y = fractextStresstextStrain = fracF / ADelta L / L implies Delta L = fracF LA Y$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} \implies \Delta L = \frac{F L}{A Y}$ Area of cross-section of wire with diameter d$d$ is: A = fracpi d^24 implies Delta L = frac4 F Lpi d^2 Y$A = \frac{\pi d^2}{4} \implies \Delta L = \frac{4 F L}{\pi d^2 Y}$ ### Core Logic Since both wires are made of the same material ($ ### Core Logic Since both wires are made of the same material ($Y_A = Y_B) and stretched with the same force ($) and stretched with the same force ($F_A = F_B): $): $Delta L propto fracLd^2$\Delta L \propto \frac{L}{d^2}$ Set up the ratio for wires A and B: $ Set up the ratio for wires A and B: $fracDelta L_ADelta L_B = left( fracL_AL_B right) times left( fracd_Bd_A right)^2$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{L_A}{L_B} \right) \times \left( \frac{d_B}{d_A} \right)^2$ ### Step 1: Substitute Given Ratios Substitute the ratios $ ### Step 1: Substitute Given Ratios Substitute the ratios $\frac{L_A}{L_B} = \frac{1}{3} and $ and $\frac{d_A}{d_B} = 2 \implies \frac{d_B}{d_A} = \frac{1}{2}: $: $fracDelta L_ADelta L_B = left( frac13 right) times left( frac12 right)^2 = frac13 times frac14 = frac112$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{1}{3} \right) \times \left( \frac{1}{2} \right)^2 = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$ Therefore, the ratio of their elongations is $ Therefore, the ratio of their elongations is $1:12. ### Pattern Recognition Sees: Same material, same stretching force. Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $. ### Pattern Recognition Sees: Same material, same stretching force. Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $(1/3) / (2^2) = 1/12$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula G = fracsigma_textsheartheta = fracFA cdot theta$G = \frac{\sigma_{\text{shear}}}{\theta} = \frac{F}{A \cdot \theta}$ where A = pi r^2$A = \pi r^2$ is the cross-sectional area. ### Core Logic Rearranging the formula for theta$\theta$: theta = fracFA cdot G = fracFpi r^2 cdot G$\theta = \frac{F}{A \cdot G} = \frac{F}{\pi r^2 \cdot G}$ Substituting the values: F = 10^5text N$F = 10^5\text{ N}$ r = 4text cm = 4 times 10^-2text m$r = 4\text{ cm} = 4 \times 10^{-2}\text{ m}$ G = 10^10text N/m^2$G = 10^{10}\text{ N/m}^2$ ### Step 1: Calculate Angular Displacement theta = frac10^5pi times (4 times 10^-2)^2 times 10^10$\theta = \frac{10^5}{\pi \times (4 \times 10^{-2})^2 \times 10^{10}}$ theta = frac10^5pi times 16 times 10^-4 times 10^10 = frac10^516pi times 10^6$\theta = \frac{10^5}{\pi \times 16 \times 10^{-4} \times 10^{10}} = \frac{10^5}{16\pi \times 10^6}$ theta = frac1160pitext radians$\theta = \frac{1}{160\pi}\text{ radians}$ ### Pattern Recognition Shear strain definition is straightforwardly matching the geometry angle of shift. Avoid radius-to-diameter or unit mismatches. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula Shear modulus definition: eta = fracsigmatheta = fracFA cdot theta$\eta = \frac{\sigma}{\theta} = \frac{F}{A \cdot \theta}$ where A = l cdot d$A = l \cdot d$ is the shearing face surface cross-area block configuration. ### Core Logic We are given that: * theta_2 = 2theta_1$\theta_2 = 2\theta_1$ * Shearing forces are identical (F_1 = F_2 = F$F_1 = F_2 = F$). * Cross section area parameter tracking: A_1 = l cdot d_1$A_1 = l \cdot d_1$, A_2 = l cdot d_2 = l cdot (2d_1) = 2A_1$A_2 = l \cdot d_2 = l \cdot (2d_1) = 2A_1$.

### Step 1: Link Modulus Terms Express deformation angles: theta_1 = fracFl cdot d_1 cdot eta_1$\theta_1 = \frac{F}{l \cdot d_1 \cdot \eta_1}$ theta_2 = fracFl cdot d_2 cdot eta_2 = fracFl cdot (2d_1) cdot eta_2$\theta_2 = \frac{F}{l \cdot d_2 \cdot \eta_2} = \frac{F}{l \cdot (2d_1) \cdot \eta_2}$ Substitute these into the relation theta_2 = 2theta_1$\theta_2 = 2\theta_1$: fracF2l d_1 eta_2 = 2 cdot left( fracFl d_1 eta_1 ight)$\frac{F}{2l d_1 \eta_2} = 2 \cdot \left( \frac{F}{l d_1 \eta_1} ight)$ frac12eta_2 = frac2eta_1 implies 4eta_2 = eta_1 implies eta_2 = fraceta_14$\frac{1}{2\eta_2} = \frac{2}{\eta_1} implies 4\eta_2 = \eta_1 implies \eta_2 = \frac{\eta_1}{4}$ ### Step 2: Solve for x Given eta_1 = 4 times 10^9mathrm~N/m^2$\eta_1 = 4 \times 10^{9}\mathrm{~N/m}^2$: eta_2 = frac4 times 10^94 = 1 times 10^9mathrm~N/m^2$\eta_2 = \frac{4 \times 10^9}{4} = 1 \times 10^{9}\mathrm{~N/m}^2$ Thus, x = 1$x = 1$. ### Pattern Recognition Be careful when identifying the dimensions of the shear face area. Doubling the slab width doubles the base support area resisting deformation, reducing stress maps under equivalent lateral loading vectors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids