Calculate the pressure due to a water column of height 120 m. (Take $g = 10 m s^{- 2}$ and density of water $= 103 k g m^{- 3}$ )

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

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12 \times 10^{5} P a

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

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

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Solution

The correct option is A $12 \times 10^{5} N m^{- 2}$
Given height of water- column = 120 m
$g = 10 m s^{- 2}$
Density of water $= 103 k g m^{- 3}$
Pressure $= h ρ g = 120 \times 10^{3} \times 10 = 12 \times 10^{5} N m^{- 2}$

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10 m

. Take density of water

= 1000 k g

m^{- 3}

g = 10 m

s^{- 2}

. the answer is

p \times 10^{5} P a

then p =

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and

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Calculate the pressure due to a water column of height 120 m. (Take g=10ms−2 and density of water =103 kg m−3)

The correct option is A 12×105 N m−2Given height of water- column = 120 m g=10 m s−2 Density of water =103 kg m−3 Pressure =h ρ g=120×103×10=12×105 N m−2

Calculate the pressure due to a water column of height 120 m. (Take $g = 10 m s^{- 2}$ and density of water $= 103 k g m^{- 3}$ )

The correct option is A $12 \times 10^{5} N m^{- 2}$
Given height of water- column = 120 m
$g = 10 m s^{- 2}$
Density of water $= 103 k g m^{- 3}$
Pressure $= h ρ g = 120 \times 10^{3} \times 10 = 12 \times 10^{5} N m^{- 2}$