A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is :

Question

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is : (A) 425 kg m-3 (B) 800 kg m-3 (C) 928 kg m-3 (D) 650 kg m-3

Tardigrade Admin · Accepted Answer

h text oil ρ text oil g=h text water ρ text water g 140 × ρ text oil =130 × ρ text water ρ text oil = (13/14) × 1000 kg / m 3 ρ text oil =928 kg m -3

$425 k g m^{- 3}$

$800 k g m^{- 3}$

$928 k g m^{- 3}$

$650 k g m^{- 3}$

$h_{oil} ρ_{oil} g = h_{water} ρ_{water} g$
$140 \times ρ_{oil} = 130 \times ρ_{water}$
$ρ_{oil} = \frac{13}{14} \times 1000 k g / m^{3}$
$ρ_{oil} = 928 k g m^{- 3}$

425kgm−3

800kgm−3

928kgm−3

650kgm−3

hoil ​ρoil ​g=hwater ​ρwater ​g 140×ρoil ​=130×ρwater ​ ρoil ​=1413​×1000kg/m3 ρoil ​=928kgm−3

$425 k g m^{- 3}$

$800 k g m^{- 3}$

$928 k g m^{- 3}$

$650 k g m^{- 3}$

$h_{oil} ρ_{oil} g = h_{water} ρ_{water} g$
$140 \times ρ_{oil} = 130 \times ρ_{water}$
$ρ_{oil} = \frac{13}{14} \times 1000 k g / m^{3}$
$ρ_{oil} = 928 k g m^{- 3}$