A capillary tube of diameter 0.4mm is dipped in a beaker containing mercury. If density of mercury is $13.6 \times 10^{3} kg / m^{3},$
surface tension 0.49Nm^-1, angle of contact 135⁰ the depression of the meniscus in the capillary tune will be

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0.026m

0.013m

0.0026m

0.13m

answer is C.

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Detailed Solution

The formula for the capillary tube's level's height is,

$h = \frac{2 Tcosθ}{rρg}$

substituting the values,

$h = \frac{2 \times 0.49 \times \cos 135}{13.6 \times 10^{3} \times \frac{0.4}{2} \times 10^{- 3} \times 9.8}$

$h = \frac{(2 \times 0.49) \times (0.707126.656)}{26.656} = 0.026 m$

Hence the correct answer is 0.026m

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