Question

Mercury is poured into a uniform vertical U-tube and water is poured above it. The level of water is the same in both arms. A piece of wood is dropped into one arm and some water equal in weight to the piece of wood is added to the other. Then, consider the statements:

(I) The level of mercury will be same in both the arms.

(II) The level of water will be same in both the arms.

• A. Both I and II are true
• B. I is true but II is false
• C. I is false but II is true
• D. Both I and II are false.

Answer 1

Answer: (A)

Both I and II are true

Answer 2

Let $W_{wood}$ be the weight of the piece of wood. Let $A$ be the cross-sectional area of the U-tube. Let $h$ be the initial height of water above the mercury in both arms.

Statement (I): The level of mercury will be same in both the arms.

In arm 1, the wood of weight $W_{wood}$ is added. The wood floats, so the buoyant force equals its weight. This means the wood displaces a volume of water $V_{sub}$ such that $\rho_{water} g V_{sub} = W_{wood}$. This weight $W_{wood}$ is effectively transmitted to the mercury interface, increasing the pressure by $\Delta P_1 = W_{wood}/A$. The pressure at the mercury interface in arm 1 is $P_1' = P_{atm} + \rho_{water} g h + \Delta P_1$.

In arm 2, water equal in weight to the piece of wood ($W_{water\_added} = W_{wood}$) is added. This increases the height of the water column by $\Delta h_2$ such that $\rho_{water} g A \Delta h_2 = W_{wood}$. The pressure at the mercury interface in arm 2 is $P_2' = P_{atm} + \rho_{water} g (h + \Delta h_2) = P_{atm} + \rho_{water} g h + \rho_{water} g \Delta h_2$.

Comparing the pressure increases: $\Delta P_1 = W_{wood}/A$ $\rho_{water} g \Delta h_2 = W_{wood}/A$ So, $\Delta P_1 = \rho_{water} g \Delta h_2$. This means $P_1' = P_2'$. Therefore, the level of mercury will be the same in both arms. Statement (I) is true.

Statement (II): The level of water will be same in both the arms.

The volume of water displaced by the wood is $V_{sub} = W_{wood} / (\rho_{water} g)$. The volume of added water in arm 2 is $V_{added\_water} = W_{wood} / (\rho_{water} g)$. Thus, $V_{sub} = V_{added\_water}$.

The initial volume of water in each arm is $V_{water\_initial} = A \times h$.

In arm 1, the total volume of water above the mercury is $V_{water\_initial} + V_{sub}$. The new water level in arm 1, $h_1$, is such that $A \times h_1 = V_{water\_initial} + V_{sub}$. $h_1 = (A \times h + V_{sub}) / A = h + V_{sub}/A$.

In arm 2, the total volume of water above the mercury is $V_{water\_initial} + V_{added\_water}$. The new water level in arm 2, $h_2$, is such that $A \times h_2 = V_{water\_initial} + V_{added\_water}$. $h_2 = (A \times h + V_{added\_water}) / A = h + V_{added\_water}/A$.

Since $V_{sub} = V_{added\_water}$, we have $h_1 = h_2$. Therefore, the level of water will be the same in both the arms. Statement (II) is true.

Both statements (I) and (II) are true.