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Question: The pressure of water on the ground floor is \[4 \times {10^4}\]Pa and on the first floor is \[{10^4...

The pressure of water on the ground floor is 4×1044 \times {10^4}Pa and on the first floor is 104{10^4}Pa. Find the height of the first floor.

Explanation

Solution

Here we use the formula of Hydrostatic Pressure and write the difference between the pressure of water on first floor and pressure on ground floor with the help of general formula of pressure where we assume the density of water as 103kgm3{10^3}\dfrac{{kg}}{{{m^3}}}and acceleration due to gravity as 10ms210\dfrac{m}{{{s^2}}}.

  • Pressure is defined as the force applied by any object. Here the formula for pressure is given by P=ρghP = \rho gh, where ρ\rho is the density of water and g is the acceleration due to gravity and h is the height from the ground level.

Complete step-by-step answer:
We are given that the pressure of water on the ground floor is 4×1044 \times {10^4}Pa and on the first floor is 104{10^4}Pa.
Let us denote the two pressures by two variables.
Let pressure of water on ground floor be denoted by P1{P_1}and the height of ground floor be h1{h_1}
P1=4×104{P_1} = 4 \times {10^4}Pa
Let pressure of water on first floor be denoted by P2{P_2} and the height of first floor be h2{h_2}
P2=104{P_2} = {10^4}Pa
We write the difference between the pressures P1{P_1}and P2{P_2}using the formula of Pressure P=ρghP = \rho gh.
⇒P1−P2=ρgh1−ρgh2\Rightarrow {P_1} - {P_2} = \rho g{h_1} - \rho g{h_2}
Since, we know the ground floor is at height 0 meters above the ground
h1=0{h_1} = 0
⇒P1−P2=−ρgh2\Rightarrow {P_1} - {P_2} = - \rho g{h_2} … (1)
Now substituting the values in equation (1)
⇒4×104−104=−(103)(10)h2\Rightarrow 4 \times {10^4} - {10^4} = - ({10^3})(10){h_2}
We can take 104{10^4}common in LHS of the equation and use the property am.an=am+n{a^m}.{a^n} = {a^{m + n}} in RHS of the equation.
⇒104(4−1)=−(103+1)h2\Rightarrow {10^4}(4 - 1) = - ({10^{3 + 1}}){h_2}
⇒3×104=−104×h2\Rightarrow 3 \times {10^4} = - {10^4} \times {h_2}
Divide both sides of the equation by 104{10^4}
⇒3×104104=−104×h2104\Rightarrow \dfrac{{3 \times {{10}^4}}}{{{{10}^4}}} = \dfrac{{ - {{10}^4} \times {h_2}}}{{{{10}^4}}}
Cancel the same terms from numerator and denominator on both sides of the equation.
⇒3=−h2\Rightarrow 3 = - {h_2}
Since we know height cannot be negative because it is a length measure.

So, the height of the first floor is 3 meters.

Note: Students should always write the unit of length along with the value obtained in the end. Also, keep in mind the unit of length should be the same as the unit used in density and acceleration due to gravity.