A body is immersed in the water. The depth of water from the surface is $h$ . Pressure acting on the top surface of the body is $P_{1}$ . If the depth is doubled and the density of the water is tripled, then the pressure becomes $P_{2}$ . What will be the ratio of $P_{1}$ and $P_{2}$ ?
Take atmospheric pressure to be $P_{o}$ .

\frac{1}{2}

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\frac{1}{3}

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\frac{1}{6}

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\frac{P_{o} + ρ g h}{P_{o} + 6 ρ g h}

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Solution

The correct option is D $\frac{P_{o} + ρ g h}{P_{o} + 6 ρ g h}$
Case 1:
Pressure at $h$ depth from the surface is:
$P_{1} = P_{o} + ρ g h$

Case 2:
If we double the depth and triple the density, then new pressure becomes:
$P_{2} = P_{o} + 3 ρ \times g \times 2 h$
$P_{2} = P_{a t m} + 6 ρ g h$

Hence, the ratio of $P_{1}$ and $P_{2}$ is:
$\frac{P_{1}}{P_{2}} = \frac{P_{o} + ρ g h}{P_{o} + 6 ρ g h}$

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A body is immersed in the water. The depth of water from the surface is h. Pressure acting on the top surface of the body is P1. If the depth is doubled and the density of the water is tripled, then the pressure becomes P2. What will be the ratio of P1 and P2? Take atmospheric pressure to be Po.

The correct option is D Po+ρghPo+6ρghCase 1: Pressure at h depth from the surface is: P1=Po+ρgh Case 2: If we double the depth and triple the density, then new pressure becomes: P2=Po+3ρ×g×2h P2=Patm+6ρgh Hence, the ratio of P1 and P2 is: P1P2=Po+ρghPo+6ρgh