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 the bottom as shown in the figure

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 the bottom as shown in the figure Question

(A) P1=P2=P3 (B) P1 < P2 < P3 (C) P1>P2>P3 (D) P2=P3≠ P1

Tardigrade Admin · Accepted Answer

According to hydrostatic paradox A , B and C points are at same pressure.

Q. 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 the bottom as shown in the figure

$P_{1} = P_{2} = P_{3}$

$P_{1} < P_{2} < P_{3}$

$P_{1} > P_{2} > P_{3}$

$P_{2} = P_{3} \neq = P_{1}$

According to hydrostatic paradox $A$ , $B$ and $C$ points are at same pressure.

Q. 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 the bottom as shown in the figure

P1​=P2​=P3​

P1​<P2​<P3​

P1​>P2​>P3​

P2​=P3​=P1​