What is the barometric height of a liquid of density 3.4 g `cm^(-3)` at a place, where that for mercury barometer is 70 cm?

Options

A. 70 cm
B. 140 cm
C. 280 cm
D. 340 cm

Answer

To find the barometric height of a liquid with a density of 3.4 g/cm³ when the barometric height of mercury is given as 70 cm, we can use the principle of hydrostatic pressure balance. Here’s a step-by-step solution:

### Step 1: Understand the relationship between pressure and height
The pressure exerted by a column of liquid is given by the formula:
\[ P = \rho g h \]
where:
- \( P \) = pressure,
- \( \rho \) = density of the liquid,
- \( g \) = acceleration due to gravity,
- \( h \) = height of the liquid column.

### Step 2: Set up the equation for mercury
For mercury, the pressure at the height of 70 cm can be expressed as:
\[ P_{Hg} = \rho_{Hg} g h_{Hg} \]
Given that:
- \( h_{Hg} = 70 \) cm,
- \( \rho_{Hg} = 13.6 \) g/cm³ (density of mercury).

Thus, the pressure exerted by mercury is:
\[ P_{Hg} = 13.6 \, \text{g/cm}^3 \cdot g \cdot 70 \, \text{cm} \]

### Step 3: Set up the equation for the liquid
For the liquid with a density of 3.4 g/cm³, we can express the pressure as:
\[ P_{L} = \rho_{L} g h_{L} \]
where:
- \( \rho_{L} = 3.4 \) g/cm³ (density of the liquid),
- \( h_{L} \) is the height of the liquid we need to find.

### Step 4: Equate the pressures
Since the atmospheric pressure is the same for both columns of liquid, we can set the two pressures equal to each other:
\[ P_{Hg} = P_{L} \]
This gives us the equation:
\[ \rho_{Hg} g h_{Hg} = \rho_{L} g h_{L} \]

### Step 5: Simplify the equation
We can cancel \( g \) from both sides:
\[ \rho_{Hg} h_{Hg} = \rho_{L} h_{L} \]

### Step 6: Solve for \( h_{L} \)
Rearranging the equation to solve for \( h_{L} \):
\[ h_{L} = \frac{\rho_{Hg} h_{Hg}}{\rho_{L}} \]

### Step 7: Substitute the known values
Substituting the known values into the equation:
- \( \rho_{Hg} = 13.6 \) g/cm³,
- \( h_{Hg} = 70 \) cm,
- \( \rho_{L} = 3.4 \) g/cm³.

So,
\[ h_{L} = \frac{13.6 \times 70}{3.4} \]

### Step 8: Calculate \( h_{L} \)
Calculating the above expression:
\[ h_{L} = \frac{952}{3.4} \approx 280 \, \text{cm} \]

### Final Answer
The barometric height of the liquid is approximately **280 cm**.

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