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Acceleration Due To Gravity

When a dolphi...

Question

When a dolphin glides through air, it experiences an external pressure of $0.75 m$ of mercury. The absolute pressure on dolphin when it is $4 m$ below the free surface of the water is
$[Take g = 10 {m/s}^{2}]$

150 kPa

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135 kPa

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142 kPa

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40 kPa

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Solution

The correct option is C $142 kPa$
Absolute pressure is given by
$P_{abs} = P_{a t m} + P_{gauge}$

$\Rightarrow P_{atm} = (ρ g H)_{mercury} = 13600 \times 10 \times 0.75 = 102000 Pa$
$\Rightarrow P_{gauge} = (ρ g H)_{water} = 1000 \times 10 \times 4 = 40000 Pa$
So,
$P_{abs} = 142000 Pa = 142 kPa$

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Acceleration Due To Gravity

Standard XII Physics

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When a dolphin glides through air, it experiences an external pressure of 0.75 m of mercury. The absolute pressure on dolphin when it is 4 m below the free surface of the water is [Take g=10m/s2]

The correct option is C 142 kPaAbsolute pressure is given by Pabs=Patm+Pgauge ⇒Patm=(ρgH)mercury=13600×10×0.75=102000 Pa ⇒Pgauge=(ρgH)water=1000×10×4=40000 Pa So, Pabs=142000Pa=142 kPa

When a dolphin glides through air, it experiences an external pressure of $0.75 m$ of mercury. The absolute pressure on dolphin when it is $4 m$ below the free surface of the water is
$[Take g = 10 {m/s}^{2}]$

The correct option is C $142 kPa$
Absolute pressure is given by
$P_{abs} = P_{a t m} + P_{gauge}$

$\Rightarrow P_{atm} = (ρ g H)_{mercury} = 13600 \times 10 \times 0.75 = 102000 Pa$
$\Rightarrow P_{gauge} = (ρ g H)_{water} = 1000 \times 10 \times 4 = 40000 Pa$
So,
$P_{abs} = 142000 Pa = 142 kPa$