Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve which is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

distance from the bottom?

Question

Physics

Madyson

25 December, 14:54

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve which is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

distance from the bottom?

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Answers (1)

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Powell · Answer 1

In balanced position, there will be only mercury in one container upto height x + y and in other container mercury upto height y and water of height 1 metre.

So pressure by x column of mercury balances water column of 1 metre. So

x. 13.6d. g = 1 x d x g (pressure = hdg and mercury is 13.6 times denser than mercury)

x = 1 / 13.6 m

= 7.35 cm

now

x + 2y = 1 (total length of mercury column is 1 m)

.0735 + 2y = 1

y = 46.32 cm

So in one column there will be mercury upto height 7.35 + 46.32 = 53.67 cm and in other column there will be mercury upto height 46.32 and water column of 1 m height.