Please solve question 153?

Question

1262 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T07:19:36

The answer is $option (2)$ $"option (2)"$

Let the atmospheric pressure be $a = 10^{5} P a$ $a=10^5Pa$

Let the depth of the lake $= h m$ $=hm$

The density of water is $ρ = 10^{3} k g m^{- 3}$ $rho=10^3kgm^-3$

Then the pressure at half the depth of the lake is

$p_{\frac{1}{2} h} = a + ρ g \frac{h}{2} = a + \frac{1}{2} ρ g h$ $p_(1/2h)=a+rhogh/2=a+1/2rhogh$

The pressure at the bottom of the lake is

$p_{b} = a + ρ g \frac{h}{2} = a + ρ g h$ $p_(b)=a+rhogh/2=a+rhogh$

Then

$p_{\frac{1}{2} h} = \frac{2}{3} p_{b}$ $p_(1/2h)=2/3p_(b)$

$a + \frac{1}{2} ρ g h = \frac{2}{3} (a + ρ g h)$ $a+1/2rhogh=2/3(a+rhogh)$

$a + \frac{1}{2} ρ g h = \frac{2}{3} a + \frac{2}{3} ρ g h$ $a+1/2rhogh=2/3a+2/3rhogh$

$\frac{1}{3} a = (\frac{2}{3} - \frac{1}{2}) ρ g h = \frac{1}{6} ρ g h$ $1/3a=(2/3-1/2)rhogh=1/6rhogh$

$h = \frac{2 a}{ρ g} = \frac{2 \cdot 10^{5}}{10 \cdot 10^{3}} = 20 m$ $h=(2a)/(rhog)=(2*10^5)/(10*10^3)=20m$

The answer is $option (2)$ $"option (2)"$