In Q.9, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Relative density of mercury = 13.6)

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into the question.

Given that the relative density of mercury is 13.6, we can employ the principle of hydrostatics to solve this problem.

Initially, let's denote the lengths of the mercury columns in the U-shaped tube as h1 and h2 in centimeters, where h1 is the height in the arm containing water and h2 is the height in the arm containing spirit.

Now, the pressure at any point in a fluid at rest is the same horizontally and vertically. Therefore, the pressure exerted by the mercury column on both sides of the tube must be equal.

The pressure exerted by a column of fluid is given by the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the column.

Initially, the pressure exerted by the mercury column on both sides is equal, so:

ρ1 * g * h1 = ρ2 * g * h2

Given that the relative density of mercury (ρ) is 13.6, we have:

13.6 * g * h1 = 13.6 * g * h2

Since the acceleration due to gravity (g) is the same on both sides and can be canceled out:

h1 = h2

This means that initially, the levels of mercury in both arms of the U-shaped tube are equal.

Now, when 15.0 cm of water and spirit each are poured into their respective arms, the level of mercury in each arm will be pushed down by 15.0 cm. Since the levels of mercury in both arms were initially the same, they will still be the same after pouring the water and spirit.

Therefore, the difference in the levels of mercury in the two arms remains unchanged at 0 cm.