What will be the pressure in $d y n e$ ${c m}^{- 2}$ , due to a water column of height $12.5 c m$ ? ( $g = 980 c m$ $s^{- 2}$ ) (density of water $10^{3} k g$ $m^{- 3}$ ).

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Solution

Pressure at the bottom of the water collumn is
$P = P_{o} + ρ g h = 10^{6} + 1 \times 980 \times 12.5$
$= 1012250 d y n e {c m}^{- 2}$

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The maximum value of water pressure in the box in which water is filled upto a height of $4 m$ is equal to (atmospheric pressure $= 10^{5} N / m^{2},$ density of water $= 1000 k g / m^{3}$ , $g = 10 m / s^{2}$ )

At what depth below the surface of water will pressure be equal to twice the atmospheric pressure? (Atmospheric pressure = 10 $N / c m^{2}$ , Density of water = $10^{3} k g / m^{3}$ , g $= 9.8 m / s^{2}$ )

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What will be the pressure in dyne cm−2, due to a water column of height 12.5cm? (g=980cm s−2) (density of water 103kg m−3).

Pressure at the bottom of the water collumn isP=Po+ρgh=106+1×980×12.5 =1012250 dyne cm−2

The maximum value of water pressure in the box in which water is filled upto a height of 4 m is equal to (atmospheric pressure =105 N/m2, density of water =1000 kg/m3, g=10 m/s2)

What will be the pressure in $d y n e$ ${c m}^{- 2}$ , due to a water column of height $12.5 c m$ ? ( $g = 980 c m$ $s^{- 2}$ ) (density of water $10^{3} k g$ $m^{- 3}$ ).

Pressure at the bottom of the water collumn is
$P = P_{o} + ρ g h = 10^{6} + 1 \times 980 \times 12.5$
$= 1012250 d y n e {c m}^{- 2}$

The maximum value of water pressure in the box in which water is filled upto a height of $4 m$ is equal to (atmospheric pressure $= 10^{5} N / m^{2},$ density of water $= 1000 k g / m^{3}$ , $g = 10 m / s^{2}$ )