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:

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

Tardigrade Admin · Accepted Answer

Pressure at the bottom P=h1 d1 g +h2 d2 g =(3 × 10-2 × 1.2 × 103 × g)+(10 × 10-2 × 103 g) =13.6 × 10-2 × 103 g=136 g =136 (g/13.6 × 103 g)=10-2 m =1 cm of Hg

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

1 cm of Hg

5 cm of Hg

13 cm of Hg

15 cm of Hg

Pressure at the bottom $P = h_{1} d_{1} g + h_{2} d_{2} g$
$= (3 \times 1 0^{- 2} \times 1.2 \times 1 0^{3} \times g) + (10 \times 1 0^{- 2} \times 1 0^{3} g)$
$= 13.6 \times 1 0^{- 2} \times 1 0^{3} g = 136 g$
$= 136 \frac{g}{13.6 \times 1 0 ^{3} g} = 1 0^{- 2} m$
$= 1 c m$ of $H g$

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