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A bucket has oil of relative density 1.2 upto a height of 3 cm. On top of the oil there is a water column of height 10 cm.If the relative density of mercury is 13.6, the pressure at the bottom of the beaker is:

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1 cm of Hg

5cm of Hg

13 cm of Hg

15 cm of Hg

answer is A.

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Detailed Solution

$P_{H g} = P_{w} + P_{o} h_{H g} ρ_{H g} g = h_{w} ρ_{w} g + h_{0} ρ_{0} g h_{H g} (13.6) = 10 (1) + 3 (1.2) h_{H g} = 1 c m o f H g$

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