### 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$\Delta P = \rho g h = 10^3 \cdot 10 \cdot 2500 = 2.5 \times 10^7 \mathrm{~N\cdot m^{-2}}$ [cite: 62, 672] 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\%$\frac{\Delta V}{V} \times 100 = \frac{\Delta P}{B} \times 100 = \frac{2.5 \times 10^7}{2 \times 10^9} \times 100\% = 1.25\%$ [cite: 672, 673] ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Mechanical Properties of Fluids