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Question

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


A

PA = PB = PC

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B

PA< PB< PC

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C

PA> PB> PC

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D

PB = PC PA

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Solution

The correct option is A

PA = PB = PC


"If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without getting diminished in magnitude”

At the same depth inside a liquid, pressure is same. A block floating will not change this affect

ThereforePA=PB=PC


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