996sorry the answer must be 2/1
An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/centimeter3) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density 4.0 grams/centimeter3 is added to one arm until a layer 5 centimeters high forms, as shown in the figure above. What is the ratio h2/h1 of the heights of the liquid in the two arms?
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5 Answers
996Is this h2/h1=4/1 the answer ?
- Swarna Kamal Dhyawala sorry the answer must be 8/5Upvote·0· Reply ·2013-01-17 08:35:13
383Let the cross-sectional area of the U-tube =a
for initial case mass of liquid in both side = (20a)gm
total mass for initial case =(40a) gm
for the 2nd case 5a cm3 liquid added
mass of this liquid = (20a) gm
total mass (40+20)a gm
that means (30a) gm each side
then h2=30cm
and
h1=10+5=15 cm
h2h1=21
[for the whole case liquid of bottom portion is negligible]
Jeet Sen Sharma same here.....
is dis d right method..??