Question

If the pressure at half depth of a lake is equal to $\dfrac{2}{3}$ pressure at the bottom of the lake then what is the depth of the lake?

Answer 1

Hint:Given the pressure at half depth of a lake is equal to $\dfrac{2}{3}$ pressure at the bottom of the lake and we have to find the depth of the lake. We find the depth with the formula $P={{P}_{0}}+pgh$. By substituting the values in the formula according to the question and solving it, we are able to get the desirable value.

Formula Used:
To find the depth of lake, we use the formula $P={{P}_{0}}+pgh$
Where P is the pressure at depth below the surface.
${{P}_{0}}$ is the pressure at the surface.
$p$ is the density of the liquid.
G is the gravitational constant.
And h is the depth below the surface at which the pressure is measured.

Complete step by step solution:
We are given that the pressure at half depth of a lake is equal to $\dfrac{2}{3}$ pressure at the bottom of the lake. We know the pressure exerted by liquid is written as,
$P=h\rho g$
Then the pressure at half depth of lake = ${{P}_{0}}+pg\dfrac{h}{2}$
And the pressure exerted at the bottom of lake = $pgh+{{P}_{0}}$
Then the equation becomes,
${{P}_{0}}+pg\dfrac{h}{2}=\dfrac{2}{3}(pgh+{{P}_{0}}) \\ $
By cross multiplying the above equation, we get
$({{P}_{0}}+pg\dfrac{h}{2})3=2(pgh+{{P}_{0}})$

On opening the brackets and simplifying the above equation, we get
$3{{P}_{0}}+pg\dfrac{3}{2}=2pgh+2{{P}_{0}} \\ $
We get ${{P}_{0}}=\left( 2-\dfrac{3}{2} \right)pgh \\ $
That is ${{P}_{0}}=\dfrac{pgh}{2} \\ $
Then $h=\dfrac{2{{P}_{0}}}{pg} \\ $
We know ${{P}_{0}}={{10}^{5}}Pa \\ $
And $\rho ={{10}^{3}}kg/{{m}^{3}} \\ $
Then $h=\dfrac{2\times {{10}^{5}}Pa}{{{10}^{3}}kg/{{m}^{3}}\times 10m/{{s}^{2}}} \\ $
$\therefore h=20\,m$

Hence, depth of the lake = 20 m

Note: Remember that as the depth under the water surface increases the pressure also increases. This is due to as the depth increases, the amount of water over the swimmer increases and the pressure also increases.