To find the depth below the surface of the liquid where the pressure is equal to the pressure exerted by a column of 5m of mercury, we can follow the steps below:
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Calculate the pressure exerted by the mercury column. The formula for pressure exerted by a liquid column is given by:
\[ P = h \cdot \rho \cdot g \]
where:
- \( h \) = height of the liquid column (in meters)
- \( \rho \) = density of the liquid (in kg/m³)
- \( g \) = acceleration due to gravity (in m/s²)
The density of mercury is given as \( \rho_{Hg} = 13.6 , \text{g/cm}^3 = 13600 , \text{kg/m}^3 \).
Now, substituting \( h = 5 , m \):
\[ P_{Hg} = 5 , m \cdot 13600 , \text{kg/m}³ \cdot 10 , \text{m/s}² = 680000 , \text{Pa} \]
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Calculate the pressure exerted by the liquid with relative density 1.3 at a certain depth. The density of the liquid with a relative density of 1.3 can be calculated as:
\[ \rho_{liquid} = 1.3 \cdot 1000 , \text{kg/m}³ = 1300 , \text{kg/m}³ \]
Now, let \( d \) be the depth below the surface of the liquid where the pressure is equal to that of the mercury column. The pressure at depth \( d \) in the liquid is:
\[ P_{liquid} = d \cdot \rho_{liquid} \cdot g \]
Setting the two pressures equal to each other gives us:
\[ d \cdot 1300 , \text{kg/m}³ \cdot 10 , \text{m/s}² = 680000 , \text{Pa} \]
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Solve for \( d \):
\[ d \cdot 13000 , \text{Pa} = 680000 , \text{Pa} \]
\[ d = \frac{680000 , \text{Pa}}{13000 , \text{Pa}} \approx 52.31 , m \]
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Determine if this depth is possible in the jar. The jar is only 40m deep, so a pressure depth of 52.31m is not feasible. This means that the pressure exerted by a 5m tall mercury column is greater than the total pressure at the bottom of the 40m deep liquid column.
Thus the pressure at the bottom of the jar, which is 40m deep:
\[ P_{max} = 40 , m \cdot 1300 , \text{kg/m}³ \cdot 10 , \text{m/s}² = 520000 , \text{Pa} \]
This means that there is no depth in the fluid where the pressure equals that of 5m of mercury, as this depth would correspond to a pressure that exceeds that generated by the jar’s liquid.
Conclusion: There is no depth below the surface of the liquid in the jar where the pressure is equal to the pressure exerted by a column of 5m of mercury.
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