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an open - ended U-tube cross - sectional...

an open - ended U-tube 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 wtare - 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 `((h_(1))/(h_(2)))` of the heights of the liquid in the two arms is

A

`(15)/(14)`

B

`(35)/(33)`

C

`(7)/(6)`

D

`(5)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B

`P_(1)=p_(2)`
`P_(0)+800gxx0.1+1000g(0.29-x)=1000g(0.29+x)+p_(0)`
`0.8+2.9-10x=2.9+10x`
`x=(0.8)/(20)=0.04`
`h_(2)=0.1+0.29-x=0.35`
`h_(1)=0.1+0.29-x=0.35`
`h_(1)=0.29+x=0.33`
`(h_(2))/(h_(1))=(35)/(33)=1.06`
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