Q: An Open – ended U-tube of uniform 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 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 figure . The ratio (h_{1}/h_{2}) of the height of the liquid in the two arms is

(a) 15/14

(b) 35/33

(c) 7/6

(d) 5/4

Ans: (b)

Sol: As h_{1} + h_{2} = 0.29 x 2 + 0.1

h_{1} + h_{2} = 0.68 …(i)

P_{0} + ρ_{k} g (0.1) + ρ_{w} g (h_{1} -0.1) – ρ_{w} g h_{2} = P_{0}

ρ_{k} g (0.1) + ρ_{w} g h_{1} – ρ_{w} g x 0.1 = ρ_{w} g h_{2}

800 x 10 x 0.1 + 1000 x 10 x h_{1} – 1000 x 10 x 0.1 = 1000 x 10 x h_{2}

10000 (h_{1} -h_{2}) = 200

h_{1} – h_{2} = 0.02 …(ii)

Solving (i) & (ii) we get

h_{1} = 0.35 m and h_{2} = 0.33 m

h_{1}/h_{2} = 35/33