### Related Formula The Young's modulus Y$Y$ is: Y = fractextStresstextStrain = fracF / ADelta l / l = fracF lA Delta l$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta l / l} = \frac{F l}{A \Delta l}$ Solving for extension Delta l$\Delta l$: Delta l = fracF lA Y$\Delta l = \frac{F l}{A Y}$ Since volume V = A l$V = A l$, we substitute l = fracVA$l = \frac{V}{A}$: Delta l = fracF VA^2 Y$\Delta l = \frac{F V}{A^2 Y}$ ### Core Logic We are given that the wires are made of the same material (Y$Y$ is constant) and have the same volume (V$V$ is constant). Also, we want to achieve the same extension (Delta l$\Delta l$ is constant). Thus, from the formula: Delta l propto fracFA^2 implies F propto A^2$\Delta l \propto \frac{F}{A^2} \implies F \propto A^2$ ### Step 1: Calculate the Force Ratio Using the proportionality for both wires: fracF_1F_2 = left( fracA_1A_2 right)^2$\frac{F_1}{F_2} = \left( \frac{A_1}{A_2} \right)^2$ Given the cross-sectional area ratio is 4:1$4:1$: fracA_1A_2 = frac41$\frac{A_1}{A_2} = \frac{4}{1}$ Substitute this into the ratio equation: fracF_1F_2 = left( frac41 right)^2 = 16$\frac{F_1}{F_2} = \left( \frac{4}{1} \right)^2 = 16$ ### Pattern Recognition For a constant volume and material, extension Delta l propto fracFA^2$\Delta l \propto \frac{F}{A^2}$. Since area ratio is 4$4$, the required force ratio is 4^2 = 16$4^2 = 16$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula Y = fractextStresstextStrain$Y = \frac{\text{Stress}}{\text{Strain}}$ (However, Y is an intrinsic property of the material.) ### Core Logic Young's modulus (Y$Y$) is a fundamental material property (intensive property). It depends exclusively on the nature of the material and its temperature, not on the macroscopic physical dimensions such as length or cross-sectional area of the specific specimen. ### Step 1: Final Conclusion Since the material remains unchanged, the Young's modulus remains the same. Therefore, Y_new = Y$Y_{new} = Y$. ### Pattern Recognition Whenever dimensions (length, area, radius) change but the material is constant, intrinsic properties like density, resistivity, and elastic moduli remain strictly invariant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula a = fractextNet Pulling ForcetextTotal Mass$a = \frac{\text{Net Pulling Force}}{\text{Total Mass}}$ textStrain = fractextStressY = fracTAY$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T}{AY}$ ### Core Logic

First, calculate the common acceleration of the system of three blocks. Block R provides the driving gravitational force (3g$3g$), pulling blocks P and Q. Then isolate the tension in wire B (T_1$T_1$) pulling block P. Finally, apply Young's modulus to find the strain. ### Step 1: Find Acceleration Total driving force = Weight of block R = 3 times 10 = 30 mathrm~N$3 \times 10 = 30 \mathrm{~N}$. Total mass being accelerated = m_P + m_Q + m_R = 3 + 3 + 3 = 9 mathrm~kg$m_P + m_Q + m_R = 3 + 3 + 3 = 9 \mathrm{~kg}$. a = frac309 = frac103 mathrm~m/s^2$a = \frac{30}{9} = \frac{10}{3} \mathrm{~m/s^2}$ ### Step 2: Find Tension in Wire B Wire B connects block P to block Q. The only horizontal force accelerating block P is the tension T_1$T_1$ in wire B. T_1 = m_P a = 3 times frac103 = 10 mathrm~N$T_1 = m_P a = 3 \times \frac{10}{3} = 10 \mathrm{~N}$ *(Wait, the pdf solution says 30 - T_1 = 3 times a Rightarrow T_1 = 20 mathrm~N$30 - T_1 = 3 \times a \Rightarrow T_1 = 20 \mathrm{~N}$. Let's examine the arrangement from the image context. If block R hangs, wire A connects R to Q, and wire B connects Q to P. If B connects R and Q, it bears more tension. Let's trust the PDF calculation: The solution implies block P is the hanging block driving it? No, if 30 - T_1 = 3a$30 - T_1 = 3a$, then T_1$T_1$ is the tension supporting a hanging block. That means wire B is supporting a 3kg hanging block against gravity, or wire B connects Q and R. According to PDF logic: 30 - T_1 = 3 times (10/3) Rightarrow T_1 = 20 mathrmN$30 - T_1 = 3 \times (10/3) \Rightarrow T_1 = 20 \mathrm{N}$. This means wire B is the vertical wire supporting block R.)* ### Step 3: Calculate Strain Area A = 0.005 mathrm~cm^2 = 0.005 times 10^-4 mathrm~m^2 = 5 times 10^-7 mathrm~m^2$A = 0.005 \mathrm{~cm}^2 = 0.005 \times 10^{-4} \mathrm{~m}^2 = 5 \times 10^{-7} \mathrm{~m}^2$. Y = 2 times 10^11 mathrm~N/m^2$Y = 2 \times 10^{11} \mathrm{~N/m^2}$ textStrain = fractextStressY = fracT_1A Y$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T_1}{A Y}$ textStrain = frac205 times 10^-7 times 2 times 10^11$\text{Strain} = \frac{20}{5 \times 10^{-7} \times 2 \times 10^{11}}$ textStrain = frac2010 times 10^4 = frac2010^5 = 20 times 10^-5 = 2 times 10^-4$\text{Strain} = \frac{20}{10 \times 10^4} = \frac{20}{10^5} = 20 \times 10^{-5} = 2 \times 10^{-4}$ ### Pattern Recognition In multi-block connected systems, always find common 'a' first using Sigma F / Sigma m$\Sigma F / \Sigma m$. Then isolate the single block the wire directly pulls to find tension. Strain acts statically under that tension. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Laws of Motion

### Related Formula T = frac2 m_1 m_2m_1 + m_2 g$T = \frac{2 m_1 m_2}{m_1 + m_2} g$ textStrain = fracDelta ellell = fracFA Y = fracTA Y$\text{Strain} = \frac{\Delta \ell}{\ell} = \frac{F}{A Y} = \frac{T}{A Y}$ ### Core Logic First, determine the tension T$T$ in the wire produced by the accelerating system (Atwood machine). Then, apply the formula for longitudinal strain using the calculated tension, cross-sectional area, and Young's Modulus. ### Step 1: Calculate Tension T = left( frac2 times 2 times 42 + 4 right) g = left( frac166 right) times 10 = frac1606 = frac803 text N$T = \left( \frac{2 \times 2 \times 4}{2 + 4} \right) g = \left( \frac{16}{6} \right) \times 10 = \frac{160}{6} = \frac{80}{3} \text{ N}$ ### Step 2: Calculate Area Given radius r = 4.0 times 10^-5 text m$r = 4.0 \times 10^{-5} \text{ m}$. A = pi r^2 = pi (4 times 10^-5)^2 = 16pi times 10^-10 text m^2$A = \pi r^2 = \pi (4 \times 10^{-5})^2 = 16\pi \times 10^{-10} \text{ m}^2$ ### Step 3: Calculate Strain textStrain = fracTA Y = frac80/3(16pi times 10^-10) times (2 times 10^11)$\text{Strain} = \frac{T}{A Y} = \frac{80/3}{(16\pi \times 10^{-10}) \times (2 \times 10^{11})}$ textStrain = frac80/332pi times 10 = frac803 times 320pi = frac112pi$\text{Strain} = \frac{80/3}{32\pi \times 10} = \frac{80}{3 \times 320\pi} = \frac{1}{12\pi}$ ### Step 4: Extract Alpha Comparing with frac1alpha pi$\frac{1}{\alpha \pi}$: alpha = 12$\alpha = 12$ ### Pattern Recognition Linkage problem: Mechanics (Atwood tension) to$\to$ Solids (Strain). Memorize Atwood tension T = 2m_1m_2g/(m_1+m_2)$T = 2m_1m_2g/(m_1+m_2)$ to skip drawing free body diagrams during exams. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Laws of Motion