Questions: Koliko je visok stupac alkohola gustoce 0,8 g / cm^3, u U-cijevi, koji je u ravnoteži sa stupcem vode od 24 cm?

Solution Steps

We need to find the height of an alcohol column in a U-tube that is in equilibrium with a water column of height 24 cm. The density of alcohol is given as \(0.8 \, \text{g/cm}^3\).

In a U-tube, the pressure at the bottom of the two columns must be equal for the system to be in equilibrium. The pressure exerted by a liquid column is given by the formula: \[ P = \rho g h \] where \(P\) is the pressure, \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the liquid column.

For the water column: \[ P_{\text{water}} = \rho_{\text{water}} g h_{\text{water}} \] For the alcohol column: \[ P_{\text{alcohol}} = \rho_{\text{alcohol}} g h_{\text{alcohol}} \] Since the pressures are equal: \[ \rho_{\text{water}} g h_{\text{water}} = \rho_{\text{alcohol}} g h_{\text{alcohol}} \]

Given:

• \(\rho_{\text{water}} = 1 \, \text{g/cm}^3\)
• \(h_{\text{water}} = 24 \, \text{cm}\)
• \(\rho_{\text{alcohol}} = 0.8 \, \text{g/cm}^3\)

Substitute these values into the equilibrium equation: \[ 1 \times g \times 24 = 0.8 \times g \times h_{\text{alcohol}} \] Cancel \(g\) from both sides: \[ 24 = 0.8 \times h_{\text{alcohol}} \] Solve for \(h_{\text{alcohol}}\): \[ h_{\text{alcohol}} = \frac{24}{0.8} = 30 \, \text{cm} \]

Final Answer