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.
Figure
(a) P1 = P2 = P3
(b) P1 < P2 < P3
(c) P1 > P2 > P3
(d) P2 = P3 ≠ P1

Answer 1

A wooden object floats in water kept in a beaker. The object is near a side of the beaker. Let $P_1, P_2, P_3$ be the pressures at the three points A, B and C of bottom, then $P_1 = P_2 = P_3$

Explanation:

As per the given problem, all three points ($P_1, P_2\, and \,P_3$) are at the same level.

As per the Pascal’s law pressure applied to incompressible fluid will transmit that pressure equally in all directions. Here water is incompressible fluid and the pressure transmitted throughout the fluid such that the same change occurs everywhere in the fluid.  

Hence pressure at the three points is same $(P_1 = P_2 = P_3)$ as the level of all the points are same if there is no acceleration.