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

0.01 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.1 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.01 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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)

Suggest Corrections

Similar questions

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

Calculate the pressure at a depth of $100 m$ below the surface of the sea. The density of sea water is $1024 \frac{k g}{m^{3}}$ .

= 1000 k g m^{- 3}, g = 10 m s^{- 2})

60 m

kPa

1, 025 {kg m}^{- 3}

1 kPa = 1000 Pa

g = 10 {m s}^{- 2}

Calculate the pressure at a depth of $50 m$ below the surface of the sea. The density of sea water is $1024 k g / m^{3}$ .

( Accelaration due to gravity = $10 m / s^{2}$ )

Join BYJU'S Learning Program