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Pascal's Law

An object of ...

Question

An object of uniform density is allowed to float in water kept in a beaker. The object has triangular cross-section as shown in the figure. If the water pressure measured at the three point A,B and C below the object are $P_{A}, P_{B}$ and $P_{C}$ respectively then:

P_{A} > P_{B} > P_{C}

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P_{A} > P_{B} < P_{C}

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P_{A} = P_{B} = P_{C}

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P_{A} = P_{C} < P_{B}

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Solution

The correct option is C $P_{A} = P_{B} = P_{C}$
As we know, that in a fluid pressure at a point is same from all side and in a stationary fluid the pressure different point at same horizontal level is also same and thus $P_{A} = P_{B} = P_{C}$

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Similar questions

Q. An object of uniform density is allowed to float in water kept in a beaker. The object has triangular cross-section as shown in the figure. If the water pressure measured at the three points

A, B

and

C

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P_{A}, P_{B}

and

P_{C}

respectively, then:

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and

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and

P_{C}

are the pressures at

A, B

and

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Q. If

A, B, C

are three events, then show that

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)

Q. If

A, B, C

are three events show that

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (b \cap C) - P (C \cap A) + P (A \cap B \cap C)

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An object of uniform density is allowed to float in water kept in a beaker. The object has triangular cross-section as shown in the figure. If the water pressure measured at the three point A,B and C below the object are PA,PB and PC respectively then:

The correct option is C PA=PB=PCAs we know, that in a fluid pressure at a point is same from all side and in a stationary fluid the pressure different point at same horizontal level is also same and thus PA=PB=PC

An object of uniform density is allowed to float in water kept in a beaker. The object has triangular cross-section as shown in the figure. If the water pressure measured at the three point A,B and C below the object are $P_{A}, P_{B}$ and $P_{C}$ respectively then:

The correct option is C $P_{A} = P_{B} = P_{C}$
As we know, that in a fluid pressure at a point is same from all side and in a stationary fluid the pressure different point at same horizontal level is also same and thus $P_{A} = P_{B} = P_{C}$