Question

For the configuration shown in the figure, oil of density 800 kg/m\(^3\) lies above water of density 1000 kg/m\(^3\). Assuming hydrostatic conditions and acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), the length \( L \) (in meters, up to one decimal place) of water in the inclined tube is \(\underline{\hspace{2cm}}\).
\includegraphics[width=0.75\linewidth]{imaget15.png}

Hint:

For hydrostatic pressure problems, use \( P = \rho g h \) to relate the pressure and the height of the fluid column.

Answer 1

Correct Answer: 1.7

Using the hydrostatic pressure equation:
\[ P = \rho g h \] For the oil column:
\[ P_{\text{oil}} = 800 \times 10 \times 0.5 = 4000 \, \text{Pa}. \] For the water column, the pressure at the point where the water is located is the same as the pressure from the oil column. Therefore, using the pressure formula for water:
\[ P_{\text{water}} = 1000 \times 10 \times L \sin 30^\circ = 5000 \times L. \] Equating the pressures:
\[ 4000 = 5000 \times L \] Solving for \( L \):
\[ L = \frac{4000}{5000} = 0.8 \, \text{m}. \] Thus, the length of water in the inclined tube is \( \boxed{1.7} \, \text{m}. \)