You visited us 1 times! Enjoying our articles? Unlock Full Access!

Pressure in Fluids

Atmospheric p...

Question

Atmospheric pressure is 1.013×10⁵Nm^-2. The pressure 100m below the surface of ocean (Specific gravity of sea water=1.03) in Nm^-2 is:-

Open in App

Solution

Density of sea water (r) is 1.03 X 10^3 kg/m3 and the atmospheric pressure is 1.01 x 10^5 N/m2.

h=100m

P(fluid)= r g h
= (1.03 x10^3 kg/m3)×(9.8 m/s2)×(100m)
= 10.094 x 10^5 N/m2

Ptotal = P(atmosphere)+ P(fluid)
= (1.013 x 10^5) +
(10. 094 x 10^5 ) Pa
= 11.107 x 10^5 Pa
=11.1 × 10^5 Pa

Suggest Corrections

Similar questions

70 \times 10^{- 3} N m^{- 1}

1.013 \times 10^{5} N m^{- 2}

1 m m

0.1 m

2 \times 10^{- 5} m .

0.75

5 \times 10^{- 2} N m^{- 1} .

1 \times 10^{5} N m^{- 2} .

1.5

1 a t m o s p h e r e = 10^{5} N m^{- 2}

30 m

1 m

1.013 \times 10^{5} {N m}^{- 2}

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

4 m

2 m

P_{a t m} = 1.013 \times 10^{5} N / m^{2}

ρ_{w}

1000 k g / m^{3}

(g) = 9.8 m s^{- 1}

Join BYJU'S Learning Program