Question

A wooden object floats in water kept in a beaker. The object is near a side of the beaker. Let P₁, P₂, P₃ be the pressures at the three points A, B and C of bottom as shown in the figure.

Options

• P₁ = P₂ = P₃

•  P₁ < P₂ < P₃

• P₁ > P₂ > P₃

• P₂ = P₃ ≠ P₁

Answer 1

P₁ = P₂ = P₃

Explanation:

The beaker's points A, B, and C are all horizontally level. The fact that

P = ρgh

where ρ is the liquid's density, g is the acceleration due to gravity, and h is the height.

As a result, the pressure in the beaker will be the same at each of the three points (A, B, and C) mentioned.

So, P₁ = P₂ = P₃