Two vessels have different base area. They are filled with water to the same height. If the amount of water in one be $4$ times that in the other, then the ratio of pressure on their bottom will be :

16 : 1

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8 : 1

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4 : 1

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1 : 1

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Solution

The correct option is D $1 : 1$
We know Pressure at depth 'H'
$P = P_{0} + ρ g H$
Since 'H' for both beaker are same
So ratio is
$1 : 1$
$Answer = (D) 1 : 1$

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Two vessels have different base area. They are filled with water to the same height. If the amount of water in one be 4 times that in the other, then the ratio of pressure on their bottom will be :

The correct option is D 1:1We know Pressure at depth 'H'P=P0+ρgHSince 'H' for both beaker are sameSo ratio is1:1 Answer =(D)1:1

Two vessels have different base area. They are filled with water to the same height. If the amount of water in one be $4$ times that in the other, then the ratio of pressure on their bottom will be :

The correct option is D $1 : 1$
We know Pressure at depth 'H'
$P = P_{0} + ρ g H$
Since 'H' for both beaker are same
So ratio is
$1 : 1$
$Answer = (D) 1 : 1$