At a depth of $1000$ m is an ocean the force acting on the window of area $20$ cm $\times 20$ cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure is [Given, density of seawater is $1.03 \times 10^{3} k g m^{- 3}, g = 10 m s^{- 2}$ ].

3.2 \times 10^{8}

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4.12 \times 10^{5}

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8.3 \times 10^{2}

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3.1 \times 10^{- 5}

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Solution

The correct option is B $4.12 \times 10^{5}$ N

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Similar questions

50 cm \times 50 cm

2000 m

= 10^{3} {kg/m}^{3}, g = 10 {m/s}^{2}

2 km

25 cm \times 20 cm

1030 {kg m}^{- 3}, g = 10 {m/s}^{2})

1.03 \times 10^{3} k g m^{- 3}, g = 10 m s^{- 2}

1.03 \times 10^{3}

k g m^{- 3}

g = 10 m s^{- 2} .

The density of seawater is $1024 k g / m^{3}$ . The pressure at a depth of $5 m$ below the surface of the sea is___ $N / m^{2}$

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