Answer:
The gate will open if the height of water is equal to or more than 0.337m.
Step-by-step explanation:
From the diagram attached, (as seen from the reference question found on google)
The forces are given as
Force on OA

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OA part of the door which is calculated as follows:
The Force is given as
![F_1=0.6\rho g[h+0.3]](https://img.qammunity.org/2021/formulas/physics/college/oxe0242dken4w9ak32n52ulpfkt5q8m8jc.png)
Force on OB

Here
- ρ is the density of water.
- g is the gravitational acceleration constant
is the equivalent height given as

is the area of the OB part of the door which is calculated as follows:

The Force is given as
![F_2=0.4\rho g[h+0.8]](https://img.qammunity.org/2021/formulas/physics/college/tsya5dh5pd9rzxkao6rq77tbpacfux32xb.png)
Now the moment arms are given as
![\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+((1)/(12)* 0.6^3 * 1)/(0.6 *[h+0.3])\\\bar{y}_a=h+0.3+(0.03)/(h+0.3)](https://img.qammunity.org/2021/formulas/physics/college/s58ukml7yf9bmgwgrkcr3vv8g6z96wgvn2.png)
![\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+((1)/(12)* 0.4^3 * 1)/(0.4 *[h+0.8])\\\bar{y}_b=h+0.8+(0.0133)/(h+0.8)](https://img.qammunity.org/2021/formulas/physics/college/kc1igmsv9dpzsxafotuvli1wadruhlps2u.png)
Taking moment about the point O as zero
/(h+0.3))=0.4\rho g[h+0.8](0.2+(0.0133)/(h+0.8))\\0.6[h+0.3](0.3-(0.03)/(h+0.3))=0.4[h+0.8](0.2+(0.0133)/(h+0.8))\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m](https://img.qammunity.org/2021/formulas/physics/college/7eufjrosq9b6ler8fo23ng1lfqbyr8ggwv.png)
So the gate will open if the height of water is equal to or more than 0.337m.