Calculate the depth at which pressure on a swimmer is 100 Pa. Given, density of water = 1000 kg/m³

0.01 m

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0.1 m

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0.01 cm

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10 m

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Solution

The correct option is A
0.01 m

Given: P = 100 Pa, density of water = ρ = 1000 kg/m³, g = 9.8 m/s²

We know that, P = ρgh

So, h = $\frac{P}{ρ g}$

i.e. h = 100 $\div$ (1000 × 9.8)

h = 0.01 m (Answer)

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Similar questions

Calculate the depth at which pressure on a swimmer is 100 Pa. Given, density of water = 1000 kg/m³

6 m

= 1000 k g

m^{- 3}

g = 10 m s^{- 2}

1300 k g

m^{- 3}

104 \times 10^{5} P a

g = 10 m

s^{- 2}

1.01 \times 10^{5}

= 1000 k g / m^{3}

___ $P a$ is the pressure at a depth of $9 m$ below the surface of the water tank. The density of water is $1000 k g / m^{3}$ .

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