Question

U-shaped tube of cross sectional area of 2cm (squared) has an amount of water 9cm (cubed) of kerosene is poured in one limb & the level difference of water becomes 3.6 cm . find the volume of benzene to be poured in the other limb so that the water level in both limbs is the same ( water's density = 1000 , Benzene's density = 900)

Answer 1

U-shaped tube of cross-sectional area of $2\mathrm{cm}^2$ has a certain amount of water in it. $9\mathrm{cm}^3$ of kerosene is poured in one limb and the level difference of water becomes $3.6\mathrm{cm}$ . Find the volume of benzene to be poured in the other limb so that the water level in both limbs is the same (water's density $= 1000\mathrm{kg / m}^3$ , benzene's density $= 900\mathrm{kg / m}^3$ ).

Solution: According to the Pascal's law, pressures in communicating vessels are equal to each other:

$\rho_{1} \cdot g \cdot h_{1} = \rho_{2} \cdot g \cdot h_{2}$ where $\rho_{1}, h_{1}$ and $\rho_{2}, h_{2}$ are the corresponding densities and level differences of water and kerosene, $\mathrm{kg} / \mathrm{m}^{3}$ and cm, respectively. When we will pour such volume of the benzene in the second limb, that the water level in both limbs will be the same, we can write a new equation: $\rho_{2} \cdot g \cdot h_{2} = \rho_{3} \cdot g \cdot h_{3}$ , where $\rho_{2}, h_{2}$ and $\rho_{3}, h_{3}$ are the corresponding densities and level heights of kerosene and benzene, $\mathrm{kg} / \mathrm{m}^{3}$ and cm, respectively. Comparing these two equations, we can make a conclusion that $\rho_{1} \cdot g \cdot h_{1} = \rho_{3} \cdot g \cdot h_{3}$ ;

$$h _ {3} = \frac {\rho_ {1} \cdot h _ {1}}{\rho_ {3}} = \frac {1 0 0 0 \cdot 3 . 6}{9 0 0} = 4 \mathrm {c m};$$

Then, volume of benzene is: $V = A \cdot h_{3} = 2 \cdot 4 = 8 \, \text{cm}^{3}$ , where $A$ is the cross-sectional area, $\text{cm}^{2}$ .

Answer: $8\mathrm{cm}^3$