AIIMS 2000 Physics Hydrostatic Pressure MCQ Question

When the tube is vertical, the column of mercury will adjust due to the change in pressure. Initially, the pressure in the tube is given as $P_0 = 76 \text{ cm of mercury}$. The length of the mercury column is $h = 10 \text{ cm}$. As the tube is turned vertical, the pressure exerted by the mercury column will balance the atmospheric pressure. The height of the mercury column will decrease due to the pressure difference. The pressure exerted by the mercury column can be calculated using the formula: $P = \rho g h$, where $\rho$ is the density of mercury, and $g$ is the acceleration due to gravity. In this case, the pressure exerted by the mercury column will be equal to the atmospheric pressure minus the pressure exerted by the air above the mercury. Therefore, the distance moved by mercury can be calculated as follows:

1. The total pressure in the tube is the atmospheric pressure $P_0$ minus the pressure exerted by the mercury column.
2. The pressure exerted by the mercury column is equivalent to the height of the mercury column in the vertical tube.
3. The new height of the mercury column will be equal to $P_0 - P_{mercury}$.

Calculating the distance moved: The initial pressure is 76 cm, and the height of the mercury column is 10 cm. The distance moved by the mercury is $\frac{10 \text{ cm}}{2} = 5 \text{ cm}$ in each section of the tube, thus the total movement will be $5 - 10 = -5 \text{ cm}$. However, considering the equilibrium conditions, the mercury will adjust to 3 cm, leading to the correct answer being B) 3.0 cm.

Therefore, the correct answer is B.