Question

For the given figure, considering Pratt's model of isostatic compensation at the crust–mantle boundary, the crustal density ($\rho_1$) that explains 1.5 km deep lake is $\underline{\hspace{1cm}}$ kg/m$^3$. (Consider density of water $\rho_w = 1000$ kg/m$^3$) [round off to 2 decimal places] \begin{center} \includegraphics[width=0.5\textwidth]{05.jpeg} \end{center}

Hint:

In Pratt's isostasy, thickness is constant but density varies. Always equate column mass per unit area at the depth of compensation, carefully accounting for replacement of crust by water in lake regions.

Answer 1

Step 1: Concept of Pratt's Model.
According to Pratt's hypothesis of isostasy, all crustal columns have the same thickness, but different densities to achieve balance at the compensation depth. \[ \text{Condition: } \rho \cdot t = \text{constant for equilibrium.} \] Here, \[ t = 30 \, \text{km (crustal thickness)} \] \[ \rho_c = 2700 \, \text{kg/m}^3 \quad \text{(continental crust density)} \] \[ h = 1.5 \, \text{km (lake depth)} \] \[ \rho_w = 1000 \, \text{kg/m}^3 \]

Step 2: Mass column for reference crust (without lake).
For the normal continental crust of thickness $t$, \[ M_c = \rho_c \cdot t = 2700 \times 30 = 81000 \, \text{kg/m}^2 \] (This is the reference column mass per unit area.)

Step 3: Mass column for lake region (with reduced density $\rho_1$).
In the lake region, total column mass = \[ M_l = \rho_1 \cdot (t - h) + \rho_w \cdot h \] Here, \[ t - h = 30 - 1.5 = 28.5 \, \text{km} \] So, \[ M_l = \rho_1 \cdot 28.5 + 1000 \cdot 1.5 \]

Step 4: Apply isostatic balance condition.
For equilibrium at compensation depth: \[ M_l = M_c \] \[ \rho_1 \cdot 28.5 + 1500 = 81000 \] \[ \rho_1 \cdot 28.5 = 79500 \] \[ \rho_1 = \frac{79500}{28.5} \]

Step 5: Final computation.
\[ \rho_1 = 2789.47 \, \text{kg/m}^3 \] Wait—check again. The lake replaces crustal material, so correction is needed. The effective condition should be: \[ \rho_c \cdot t = \rho_1 \cdot t - (\rho_1 - \rho_w)h \] \[ 81000 = \rho_1 \cdot 30 - (\rho_1 - 1000)(1.5) \] \[ 81000 = 30\rho_1 - 1.5\rho_1 + 1500 \] \[ 81000 - 1500 = 28.5 \rho_1 \] \[ 79500 = 28.5 \rho_1 \] \[ \rho_1 = \frac{79500}{28.5} = 2789.47 \, \text{kg/m}^3 \]

Final Answer:
\[ \boxed{2789.47 \, \text{kg/m}^3} \]