91Ó°ÊÓ

The initial height of oil and water in both tubes is, $h=25cm$.

The specific gravity of oil is, $s_o=0.80$.

The final heights of oil and water in tubes s found by calculating the rise and fall of oil and water in tubes. The water is heavier than oil, so that water will force oil upward in the tube, and the level of water dips in the tube.

The rise of oil or fall of water level in tubes is given as:

$\begin{array}{r}{s}_{w}g(h−\mathrm{Δ}h)=s_{o}g(h+\mathrm{Δ}h)\\ s_w(h−\mathrm{Δ}h)=s_o(h+\mathrm{Δ}h)\end{array}$

Here, $s_w$ is the specific gravity of water, and its value is 1 .

Substitute all the values in the above equation, and we get,

$\text{(1)}\begin{array}{r}(25\mathrm{cm}−\mathrm{Δ}h)=\left(0.80\right)(25\mathrm{cm}+\mathrm{Δ}h)\\ \mathrm{Δ}h=4.2\mathrm{cm}\end{array}$

The final height of the water is given as:

$\begin{array}{r}{H}_w=h−\mathrm{Δ}h\\ H_w=25\mathrm{cm}−4.2\mathrm{cm}\\ H_w=20.8\mathrm{cm}\end{array}$

The final height of oil is given as:

$\begin{array}{r}{H}_0=h+\mathrm{Δ}h\\ H_0=25\mathrm{cm}+4.2\mathrm{cm}\\ H_0=29.2\mathrm{cm}\end{array}$

Therefore, the final heights of water and oil in tubes are 20.8 cm and 29.2 cm .

(i)

The height on each side will remain 25 cm for the same density of oil and water because the weight of water and oil will be the same for the same densities.

Therefore, the height on each side for the same density of oil and water is .

(ii)

The water in the left tube will force oil in the right tube with greater force, and water will reach in the oil tube, so the oil will be raised completely in its tube, and water will remain below the oil.

The height in arm A will be,

$=12.5cm$

The height in arm B will be,

$$\begin{gathered} =30cm+12.5cm \\ =32.5cm \end{gathered}$$

Therefore, the heights are 12.5 cm and 32.5 cm in arms A and B, respectively, and the water will enter in oil tube for a much small density of oil.