Question

A wooden object floats in water kept in a beaker. the object is near a side of the beaker (figure 13-Q4). Let P₁, P₂, P₃ be the pressures at the three points A, B and C of the bottom as shown in the figure.

Answer 1

Options are not given.

(a) P₁ = P₂ = P₃, (b) P₁ < P₂ < P₃, (c) P₁ > P₂ > P₃ and (d) P₂ = P₃ ≠ P₁

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Answer ⇒ Option (a).

Explanation ⇒ We know that pressure at the horizontal point in stationary liquid remains the same.

Proof ⇒ Let there are two points A and B is horizontal plane of vessel at an height of h from top of vessel.

Thus, Pa = P₀ + hρg

Similarly, Pb = P₀ + hρg

Thus, we can see that Pa = Pb.

Hence, pressure remains same at horizontal point.

Thus, Pa = Pb = Pc.

Hence, Option (a). is surely correct.

Hope it helps.