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 $800 kg m ^{-3}$ is added to the left arm until its length is $0.1 m$, as shown in the schematic figure below. The ratio $\left(\frac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is-
IIT 2020, Medium
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$h _1+ h _2=0.29 \times 2+0.1$
$h _1+ h _2=0.68$ $. . . . . . .(2)$
$\Rightarrow P _0+\rho_{ k } g (0.1)+\rho_\pi g \left( h _1-0.1\right)\left[\rho_{ k }=\text { density of kerosene } \ \rho_{ w }=\text { density of water }\right]$
$-\rho_\pi gh _2= P _0$
$\Rightarrow \rho_{ k } g (0.1)+\rho_\pi gh _1-\rho_\pi g \times(0.1)$
$=\rho_\pi gh _2$
$\Rightarrow 800 \times 10 \times 0.1+1000 \times 10 \times h _1$
$-1000 \times 10 \times 0.1=1000 \times 10 \times h _2$
$\Rightarrow 10000\left( h _1- h _2\right)=200$
$\Rightarrow h _1- h _2=0.02$ $. . . . . . .(2)$
$\Rightarrow h_1=0.35$
$\Rightarrow h_2=0.33$
So, $\frac{ h _1}{ h _2}=\frac{35}{33}$
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