An open-ended u-tube of uniform of uniform cross-sectional area contains water (density $1.0 g r a m / {c m}^{3}$ ) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density $4.0 g r a m s / {c m}^{3}$ 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?

Question

An open-ended u-tube of uniform of uniform cross-sectional area contains water (density

1.0 g r a m / {c m}^{3}

) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density

4.0 g r a m s / {c m}^{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?

Toppr Admin · Accepted Answer

H1-h2-20-20-5-45 -I- since initially water was 20 cm in each arm- Since pressure at bottom of both the limbs is equal after adding the immiscible liquid- we have5-xD7-4-xD7-g-h1-x2212-5-xD7-1-xD7-g-h2-xD7-1-xD7-g-x21D2-h2-x2212-h1-15 -II-from -I- and -II- we haveh2-30-h1-15-x21D2-h2h1-21

h1+h2=20+20+5=45 ...............(I), since initially water was 20 cm in each arm. Since pressure at bottom of both the limbs is equal after adding the immiscible liquid, we have5×4×g+(h1−5)×1×g=h2×1×g⇒h2−h1=15 ...............(II)from (I) and (II), we haveh2=30,h1=15⇒h2h1=21