Question

A U-tube with both end 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 $10mm$ above the water level on the other side. Meanwhile the water rises by $65mm$ from its original level (see diagram). The density of the oil is

• A. $425kg{m}^{-3}$
• B. $800kg{m}^{-3}$
• C. $928kg{m}^{-3}$
• D. $650kg{m}^{-3}$

Answer 1

The correct option is C $928kg{m}^{-3}$

Pressure at point B and C is same.
Let ${\rho }_o$ be the density of oil and ${\rho }_w$ be the density of water. Therefore,
${\rho }_{o}g\times 140\times {10}^{-3}={\rho }_{w}g\times 130\times {10}^{-3}$
${\rho }_o=\frac{130}{140}\times {10}^3\approx 928kg{m}^{-3}$