Question

An open-ended U-tube of uniform cross-sectional area contains water (density ${10}^{3}kg{m}^{-3}$). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800kg{m}^{-3}$ is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio (ℎ1/ℎ2) of the heights of the liquid in the two arms is

• A. $\frac{15}{14}$
• B. $\frac{35}{33}$
• C. $\frac{7}{6}$
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

$\frac{35}{33}$

Correct answer is (B) $\frac{35}{33}$

$h_1+h_2=0.29\times 2+0.1$

$h_1+h_2=0.68$..........(1)

$\Rightarrow P_0+{\rho }_{k}g\left(0.1\right)+{\rho }_{w}g(h_1-0.1)$

[${\rho }_0$= density of kerosene and ${\rho }_w$ = density of water]

$-{\rho }_{w}g{h}_2=P_0$

$\Rightarrow {\rho }_{k}g\left(0.1\right)+{\rho }_{w}g{h}_1-{\rho }_{w}g\times \left(0.1\right)$

$={\rho }_{w}g{h}_2$

$\Rightarrow 800\times 10\times 0.1+1000\times 10\times h_1$

$\Rightarrow -1000\times 10\times 0.1=1000\times 10\times h_2$

$\Rightarrow 10000(h_1-h_2)=200$

$\Rightarrow h_1-h_2=0.02$.........(2)

$h_1=0.35$

$h_2=0.33$

So, $\frac{h_1}{h_2}=\frac{35}{33}$