What would be the absolute pressure at a dept | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The absolute pressure $P$ at a depth $h$ is given by the formula: $P = P_{atm} + \rho gh$.Given values are: $h = 1 \,km = 1000 \,m$,$\rho = 10^{3}

Vedclass · Accepted Answer

$1.011 	imes 10^{7} \,N/m^{2}$

Vedclass · Answer

(A) The absolute pressure $P$ at a depth $h$ is given by the formula: $P = P_{atm} + ho gh$.Given values are: $h = 1 \,km = 1000 \,m$,$ho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,and $P_{atm} = 1.01 	imes 10^{5} \,N/m^{2}$.Calculating the gauge pressure (pressure due to water column): $P_{gauge} = ho gh = 10^{3} 	imes 10 	imes 1000 = 10^{7} \,N/m^{2}$.Now,adding the atmospheric pressure: $P = 1.01 	imes 10^{5} + 10^{7} = 0.0101 	imes 10^{7} + 10^{7} = 1.0101 	imes 10^{7} \,N/m^{2}$.Rounding to the appropriate significant figures,we get $P \approx 1.011 	imes 10^{7} \,N/m^{2}$.

What would be the absolute pressure at a depth of $1 \,km$ below the ocean surface? [Given: density of water $\rho = 10^{3} \,kg/m^{3}$,$g = 10 \,m/s^{2}$,$1 \,atm = 1.01 \times 10^{5} \,N/m^{2}$]

Similar Questions

The pressure at a point $10 \ m$ below the surface of water is (take atmospheric pressure $P_0 = 1.01 \times 10^5 \ Nm^{-2}$ and $g = 10 \ ms^{-2}$):

The heights of mercury in a barometer at the base and at the top of a mountain are $75 \, cm$ and $50 \, cm$,respectively. If the ratio of the density of mercury to the density of air is $10^4$,what is the height of the mountain?

The density of the liquid filled in the container is $900 \ kg/m^{3}$. Calculate the force exerted on the bottom of the container. (Given: $g = 10 \ m/s^{2}$) (in $N$)

The atmospheric pressure at a place is $10^5 \,Pa$. If tribromomethane (specific gravity $= 2.9$) is employed as the barometric liquid,the barometric height is .......... $m$. (in $.52$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module