An open-ended $U$-tube of uniform cross-sectional area contains water (density $1.0 $ gram/centimeter$^3$) standing initially $20$ centimeters from the bottom in each arm. An immiscible liquid of density $4.0$ grams/ centimeter $^3$ is added to one arm until a layer $5$ centimeters high forms, as shown in the figure above. What is the ratio $h_2/h_1$ of the heights of the liquid in the two arms?
Diffcult
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$h_{1}+h_{2}=20+20+5=45 \quad…………(I),$ since initially water was $20 \mathrm{cm}$ in each arm. since pressure at bottom of both the limbs is equal after adding the immiscible liquid,
we have
$5 \times 4 \times g+\left(h_{1}-5\right) \times 1 \times g=h_{2} \times 1 \times g$
$\Rightarrow h_{2}-h_{1}=15………..(I I)$
from $(I)$ and $(I I),$ we have
$h_{2}=30, h_{1}=15$
$\Rightarrow \frac{h_{2}}{h_{1}}=\frac{2}{1}$
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