### Related Formula Bulk Modulus formula: B = - fracDelta PfracDelta VV implies Delta P = B left( frac-Delta VV ight)$B = - \frac{\Delta P}{\frac{\Delta V}{V}} \implies \Delta P = B \left( \frac{-\Delta V}{V} ight)$ ### Core Logic Given specifications: - Fractional volume drop, frac-Delta VV = 0.2\% = frac0.2100 = 2 times 10^-3$\frac{-\Delta V}{V} = 0.2\% = \frac{0.2}{100} = 2 \times 10^{-3}$ - Bulk modulus value, B = 2.15 times 10^9\ mathrmN/m^2$B = 2.15 \times 10^{9}\ \mathrm{N/m^2}$ Calculating excess pressure: Delta P = 2.15 times 10^9 times 2 times 10^-3$\Delta P = 2.15 \times 10^{9} \times 2 \times 10^{-3}$ Delta P = 4.3 times 10^6\ mathrmN/m^2 = 43 times 10^5\ mathrmN/m^2$\Delta P = 4.3 \times 10^{6}\ \mathrm{N/m^2} = 43 \times 10^{5}\ \mathrm{N/m^2}$ Comparing with P times 10^5$P \times 10^{5}$, we get P = 43$P = 43$. ### Pattern Recognition Bulk modulus measures a fluid's resistance to compression. A tiny percentage reduction in volume requires a massive amount of pressure due to water's near-incompressibility. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula Bulk modulus B$B$ measures a material's resistance to uniform compression and is defined as the ratio of hydraulic pressure to volumetric strain: B = fracDelta Pleft(fracDelta VVright) implies Delta V = fracDelta P cdot VB$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} \implies \Delta V = \frac{\Delta P \cdot V}{B}$ ### Core Logic Given parameters [cite: 192, 193]: * Edge length of the cube, a = 10 text cm = 0.1 text m$a = 10 \text{ cm} = 0.1 \text{ m}$ * Initial volume, V = a^3 = (0.1)^3 = 10^-3 text m^3 = 10^6 text mm^3$V = a^3 = (0.1)^3 = 10^{-3} \text{ m}^3 = 10^6 \text{ mm}^3$ * Hydraulic pressure increase, Delta P = 7 times 10^6 text Pa$\Delta P = 7 \times 10^6 \text{ Pa}$ * Bulk Modulus, B = 1.4 times 10^11 text N/m^2$B = 1.4 \times 10^{11} \text{ N/m}^2$ Substitute these values into the volume change equation : Delta V = frac7 times 10^6 times 10^-31.4 times 10^11$\Delta V = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}}$ Delta V = frac7 times 10^31.4 times 10^11 = 5 times 10^-7 text m^3 quad text$\Delta V = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-7} \text{ m}^3 \quad \text{}$ Convert the volume contraction into textmm^3$\text{mm}^3$: Delta V = 5 times 10^-7 times (10^3)^3 text mm^3 = 5 times 10^-7 times 10^9 text mm^3 = 50 text mm^3$\Delta V = 5 \times 10^{-7} \times (10^3)^3 \text{ mm}^3 = 5 \times 10^{-7} \times 10^9 \text{ mm}^3 = 50 \text{ mm}^3$ ### Pattern Recognition Always perform unit conversions carefully at the final step to avoid handling complex decimals during calculations. Converting 1 text m^3 = 10^9 text mm^3$1 \text{ m}^3 = 10^9 \text{ mm}^3$ ensures a clean, error-free conversion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids

### Related Formula B = fracDelta Pleft(fracDelta VVright) = fracrho g hleft(fracDelta VVright)$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} = \frac{\rho g h}{\left(\frac{\Delta V}{V}\right)}$ ### Core Logic Calculate the hydrostatic pressure change at depth h = 2.5text km = 2500text m$h = 2.5\text{ km} = 2500\text{ m}$ : Delta P = rho g h = 10^3 cdot 10 cdot 2500 = 2.5 times 10^7 mathrm~Ncdot m^-2$\Delta P = \rho g h = 10^3 \cdot 10 \cdot 2500 = 2.5 \times 10^7 \mathrm{~N\cdot m^{-2}}$ Now compute the percentage fractional compression : fracDelta VV times 100 = fracDelta PB times 100 = frac2.5 times 10^72 times 10^9 times 100\% = 1.25\%$\frac{\Delta V}{V} \times 100 = \frac{\Delta P}{B} \times 100 = \frac{2.5 \times 10^7}{2 \times 10^9} \times 100\% = 1.25\%$ ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Mechanical Properties of Fluids

### Related Formula Intermolecular forces weaken as thermal agitation increases: Y = fractextStresstextStrain$Y = \frac{\text{Stress}}{\text{Strain}}$ ### Core Logic When temperature rises, the mean separation between atoms or molecules increases due to thermal expansion. This weakens the intermolecular binding forces, making the material easier to deform for the same amount of applied stress. Consequently, the value of Young's modulus decreases with a rise in temperature. ### Step 1: Final Conclusion Thus, Young's modulus of elasticity decreases with the increase in temperature. ### Pattern Recognition Temperature up rightarrow$\rightarrow$ Thermal expansion up rightarrow$\rightarrow$ Intermolecular bonds weaken rightarrow$\rightarrow$ Modulus of elasticity down. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids