Question

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

• A. 1 cm of Hg
• B. 5 cm of Hg
• C. 13 cm of Hg
• D. 15 cm of Hg

Answer 1

Answer: (A)

Pressure at the bottom $P=h_{1} d_{1} g +h_{2} d_{2} g$
$=\left(3 \times 10^{-2} \times 1.2 \times 10^{3} \times g\right)+\left(10 \times 10^{-2} \times 10^{3} g\right)$
$=13.6 \times 10^{-2} \times 10^{3} g=136\, g$
$=136 \frac{g}{13.6 \times 10^{3} g}=10^{-2} m$
$=1\, cm$ of $Hg$