Water is kept in a beaker, and a wooden object floats in it. The object is located near one side of the beaker, as shown in the figure. Let $P_1$ , $P_2$ , and $P_3$ denote the pressures at the three bottom points $A$ , $B$ , and $C$ , respectively.

Question

Water is kept in a beaker, and a wooden object floats in it. The object is located near one side of the beaker, as shown in the figure. Let

P_1

,

P_2

, and

P_3

denote the pressures at the three bottom points

A

,

B

, and

C

, respectively.
Water is kept in a beaker, and a wooden object floats in it. The object is located near one side of the beaker, as

Quantrex Academy · Accepted Answer

Correct answer: D. P_1 = P_2 = P_3

In a liquid at rest, the pressure is the same at all points on the same horizontal level. If there were a pressure difference between any two points at the same level, it would create a horizontal force, causing the liquid to flow. Since the water is in static equilibrium, no such flow occurs.

The floating wooden block displaces some water, causing the overall water level in the beaker to rise uniformly. The pressure at the bottom of the beaker depends only on the atmospheric pressure and the total height of the water column (

P = P_{atm} + \rho gh

). Because the water surface is horizontal, the depth

h

is the same for all points at the bottom.

Therefore, the pressures at points A, B, and C, which lie on the same horizontal plane at the bottom of the beaker, are equal.

Thus,

P_1 = P_2 = P_3

.

Water is kept in a beaker, and a wooden object floats in it. The...

Options

Correct answer

Step-by-step solution

More from Fluid Mechanics