A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side until water rises by 25 cm on the other side. If the density of oil is 0.8, the oil level will stand higher than the water level by  

On pouring oil on the left side water rises by 25 cm from its previous level in the right limb of U-tube, creating a difference of levels of water by 50 cm. Let h cm be the height of oil above level A in the left limb of U-tube. Equating pressures at A and B, we get

${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{B}}\text{. }$

$\Rightarrow \mathrm{h}\times {\rho }_{\text{oil }}\times \mathrm{g}=50\times {\rho }_{\text{water }}\times \mathrm{g}$

$\Rightarrow \mathrm{h}=\frac{50\times {\rho }_{\text{water }}}{{\rho }_{\text{oil }}}=\frac{50\times 1}{0.8}=62.5\mathrm{cm}$

Difference of levels of oil (C) and water (D) in the two limbs

= 62.5 – 50 = 12.5 cm