An open capillary tube is lowered in vessel with mercury. The difference between the levels of the mecury in the vessel and in the capillary tube `/_\h=4.6 mm`. What is the radius of curvature of the mercury meniscus in the capillary tube? Surface tension of mercury is `0.46N//m`, density of mercury is `13.6gm//"cc"`.

Options

A. (a)`1/340m`
B. (b)`1/680m`
C. (c)`1/1020m`
D. (d)information insufficient

Explanation

To find the radius of curvature of the mercury meniscus in the capillary tube, we can use the formula related to capillarity. The height difference (\(\Delta h\)) between the mercury levels in the vessel and the capillary tube is given by:

\[ \Delta h = \frac{2S \cos \theta}{\rho g R} \]

Where:
- \(S\) = surface tension of the liquid (mercury in this case)
- \(\theta\) = contact angle (for mercury, it is approximately 0 degrees, hence \(\cos \theta = 1\))
- \(\rho\) = density of the liquid (mercury)
- \(g\) = acceleration due to gravity
- \(R\) = radius of curvature of the meniscus

### Given Data:
- \(\Delta h = 4.6 \, \text{mm} = 0.0046 \, \text{m}\)
- \(S = 0.46 \, \text{N/m}\)
- \(\rho = 13.6 \, \text{g/cm}^3 = 13600 \, \text{kg/m}^3\)
- \(g \approx 10 \, \text{m/s}^2\)

### Step 1: Rearranging the Formula
We can rearrange the formula to solve for \(R\):

\[ R = \frac{2S \cos \theta}{\rho g \Delta h} \]

### Step 2: Substitute the Values
Substituting the known values into the equation:

\[ R = \frac{2 \times 0.46 \times 1}{13600 \times 10 \times 0.0046} \]

### Step 3: Calculate the Denominator
Calculating the denominator:

\[ \rho g \Delta h = 13600 \times 10 \times 0.0046 = 625.6 \]

### Step 4: Calculate the Numerator
Calculating the numerator:

\[ 2S \cos \theta = 2 \times 0.46 \times 1 = 0.92 \]

### Step 5: Final Calculation for R
Now substituting back into the equation for \(R\):

\[ R = \frac{0.92}{625.6} \approx 0.00147 \, \text{m} = 1.47 \, \text{mm} \]

### Conclusion
The radius of curvature of the mercury meniscus in the capillary tube is approximately \(1.47 \, \text{mm}\).

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