An open-ended U-tube of uniform cross-sectional area contains water (density 103 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 (( h 1/ h 2)) of the heights of the liquid in the two arms is

Question

An open-ended U-tube of uniform cross-sectional area contains water (density 103 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 (( 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)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2143daf2dca4cabca0a6f43b1cc49321-.png" /> 800 × 0.1+1000 × x=1000 ×(0.58-x) 2000 x=500 x=0.25 m So, h1=0.35 m , h2=0.33 m (h1/h2)=(35/33)

$\frac{15}{14}$

$\frac{35}{33}$

$\frac{7}{6}$

$\frac{5}{4}$

$800 \times 0.1 + 1000 \times x = 1000 \times (0.58 - x)$
$2000 x = 500$
$x = 0.25 m$
So, $h_{1} = 0.35 m, h_{2} = 0.33 m$
$\frac{h _{1}}{h _{2}} = \frac{35}{33}$

1415​

3335​

67​

45​

800×0.1+1000×x=1000×(0.58−x) 2000x=500 x=0.25m So, h1​=0.35m,h2​=0.33m h2​h1​​=3335​

$\frac{15}{14}$

$\frac{35}{33}$

$\frac{7}{6}$

$\frac{5}{4}$

$800 \times 0.1 + 1000 \times x = 1000 \times (0.58 - x)$
$2000 x = 500$
$x = 0.25 m$
So, $h_{1} = 0.35 m, h_{2} = 0.33 m$
$\frac{h _{1}}{h _{2}} = \frac{35}{33}$