The speed of a longitudinal wave in a metallic bar is 400 text m/s$400 \text{ m/s}$. If the density and Young's modulus of the bar material are increased by 0.5% and 1% respectively then the speed of the wave is changed approximately to ____ m/s.

### 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 textThermal Tension, T = Y A alpha Delta theta$$\text{Thermal Tension, } T = Y A \alpha \Delta \theta$$ ### Core Logic For the first cooling, the tension T$T$ generated is: T = Y A alpha Delta theta_1$$T = Y A \alpha \Delta \theta_1$$ T = Y A alpha (27 - (-43))$$T = Y A \alpha (27 - (-43))$$ T = Y A alpha (70) quad dots text(i)$$T = Y A \alpha (70) \quad \dots \text{(i)}$$ Let theta$\theta$ be the final temperature required for a tension of 1.4T$1.4T$. 1.4T = Y A alpha (27 - theta) quad dots text(ii)$$1.4T = Y A \alpha (27 - \theta) \quad \dots \text{(ii)}$$ ### Step 1: Ratio of Tensions Dividing equation (ii) by equation (i): frac1.4TT = frac27 - theta70$$\frac{1.4T}{T} = \frac{27 - \theta}{70}$$ 1.4 = frac27 - theta70$$1.4 = \frac{27 - \theta}{70}$$ 27 - theta = 1.4 times 70 = 98$$27 - \theta = 1.4 \times 70 = 98$$ theta = 27 - 98 = -71^circ textC$$\theta = 27 - 98 = -71^{\circ} \text{C}$$ ### Pattern Recognition Thermal tension strictly scales with the absolute difference in temperature from the unstressed state, Delta theta$\Delta \theta$. Multiplying tension by a factor simply scales Delta theta$\Delta \theta$ by that identical factor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Thermal Properties of Matter

### Related Formula Y = fractextStresstextStrain = fracF / ADelta ell / ell = fracF ellA Delta ell$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta \ell / \ell} = \frac{F \ell}{A \Delta \ell}$$ ### Core Logic Since the wires are stretched by the same magnitude (Delta ell_A = Delta ell_B$\Delta \ell_A = \Delta \ell_B$) under the same load (F_A = F_B$F_A = F_B$), the Young's Modulus Y$Y$ is proportional to fracellA$\frac{\ell}{A}$. ### Step 1: Setting up the Ratio fracY_AY_B = left(fracell_Aell_Bright) left(fracA_BA_Aright)$$\frac{Y_A}{Y_B} = \left(\frac{\ell_A}{\ell_B}\right) \left(\frac{A_B}{A_A}\right)$$ Substitute the given values: ell_A = 6.0 mathrm~cm$\ell_A = 6.0 \mathrm{~cm}$ ell_B = 5.4 mathrm~cm$\ell_B = 5.4 \mathrm{~cm}$ A_A = 3.0 times 10^-5 mathrm~m^2$A_A = 3.0 \times 10^{-5} \mathrm{~m}^2$ A_B = 4.5 times 10^-5 mathrm~m^2$A_B = 4.5 \times 10^{-5} \mathrm{~m}^2$ ### Step 2: Calculation fracY_AY_B = left(frac6.05.4right) left(frac4.5 times 10^-53.0 times 10^-5right)$$\frac{Y_A}{Y_B} = \left(\frac{6.0}{5.4}\right) \left(\frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}}\right)$$ = left(frac6054right) left(frac4.53.0right)$$= \left(\frac{60}{54}\right) \left(\frac{4.5}{3.0}\right)$$ = left(frac109right) left(frac1.51right) = frac159 = frac53$$= \left(\frac{10}{9}\right) \left(\frac{1.5}{1}\right) = \frac{15}{9} = \frac{5}{3}$$ ### Step 3: Finding x Given ratio is x : 3$x : 3$. fracx3 = frac53 Rightarrow x = 5$$\frac{x}{3} = \frac{5}{3} \Rightarrow x = 5$$ ### Pattern Recognition Whenever "stretched by same magnitude under same load" appears, strip out F$F$ and Delta ell$\Delta \ell$ to leave Y propto ell / A$Y \propto \ell / A$. ### 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