A brass wire of length 2m and radius 1mm at 27^circ$27^{\circ}$ C is held taut between two rigid supports. Initially it was cooled to a temperature of -43^circ$-43^{\circ}$ C creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T, is ____^circ$^{\circ}$C

### Related Formula Y = fractextStresstextStrain$$Y = \frac{\text{Stress}}{\text{Strain}}$$ ### Core Logic In a plot where Strain is on the y-axis and Stress is on the x-axis, the slope of the line equals Strain/Stress. Slope = fractextStraintextStress = frac1Y$$Slope = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}$$ Since Y = frac1Slope$Y = \frac{1}{Slope}$, the material with the largest Young's modulus Y$Y$ will correspond to the line with the minimum slope (least steep line). ### Step 1: Final Conclusion From the given graph, curve C is closest to the x-axis, meaning it has the smallest slope. Therefore, material C has the largest Young's modulus. ### Pattern Recognition Sees: "strain-stress plot" instead of standard "stress-strain plot" → The axes are swapped! Always double check axis labels. Strain on Y means Slope = 1/Y. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### 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