First slide

Fluid statics

Question

The force acting on a window of area 50 cm x 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = $10^{3} kg m^{- 3}$ , g = l0 ${ms}^{- 2}$ )

Easy

A

$5 \times 10^{5} N$
B

$25 \times 10^{5} N$
C

$5 \times 10^{6} N$
D

$25 \times 10^{6} N$

Solution

Here, h = 2000 m, $ρ = 10^{3} kg m^{- 3}, g = 10 {ms}^{- 2}$
The pressure outside the submarine is $P = P_{a} + ρgh$
where $P_{a} is the atmospheric pressure$
Pressure inside the submarine is $P_{a}$ .
Hence, net pressure acting on the window is guage pressure.
Gauge pressure, $P_{g} = P - P_{a} = ρgh$
= $10^{3} kg m^{- 3} \times 10 {ms}^{- 2} \times 2000 m = 2 \times 10^{7} Pa$
Area of a window is
A = 50 cm $\times 50 cm = 2500 \times 10^{- 4} m^{2}$
Force acting on the window is
F = $P_{g} A = 2 \times 10^{7} Pa \times 2500 \times 10^{- 4} m^{2}$
= 5 $\times 10^{6} N$

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